Maths Class 12 Probability, R.D Sharma Question Answer

Page 31.16 Ex. 31.1

Q1.

Answer :

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Consider the given events.
A = An even number on the card
B = A number more than 3 on the card

Clearly,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}

Now, A∩B=4,6,8,10∴ Required probability = PA/B=nA∩BnB=47

Q2.

Answer :

Consider the given events.
A = Both the children are girls.
B = The youngest child is a girl.
C = At least one child is a girl.

Clearly, S=B1B2, B1G2, G1B2, G1G2A=G1G2B=B1G2, G1G2 C=B1G2,G1B2,G1G2

A∩B=G1G2 and A∩C=G1G2i Required probability = PA/B=nA∩BnB=12ii Required probability = PA/C=nA∩BnC=13

Q3.

Answer :

Consider the given events
A = Numbers appearing on two dice are different
B = The sum of the numbers on two dice is 4

Clearly,
A = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 3), (2, 4), (2, 5), (2, 6),
(3, 1), (3, 2), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4, 3), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 6),
(6, 1), (6, 2), (6, 3), (6, 4), (6, 5)}
B = {(1, 3), (3, 1) and (2, 2)}

Now, A∩B=1,3 and 3,1∴ Required probability = PB/A=nA∩BnA=230=115

Q4.

Answer :

Consider the given events.
A = Getting head on third toss
B = Getting head on first two tosses

Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}

Now,A∩B=H,H,H∴ Required probability = PA/B=nA∩BnB=12

Q5.

Answer :

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and 5 on second throw

Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4),
(2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),
. .
. .
. .
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

Now, A∩B=6,5,4∴ Required probability = PA/B=nA∩BnB=16

Q6.

Answer :

Given:PB=0.5PA∩B=0.32Now, PAB=PA∩BPB⇒ PAB=0.320.5=3250=1625

Q7.

Answer :

Given:PA=0.4PB=0.3PB/A=0.5Now,PB/A=PA∩BPA⇒0.5=PA∩B0.4⇒PA∩B=0.2PA/B=PA∩BPB=0.20.3=23

Q8.

Answer :

Given:PA=13PB=15PA∪B=1130Now, PA∪B=PA+PB-PA∩B⇒1130=13+15-PA∩B⇒PA∩B=13+15-1130=10+6-1130=530=16PA/B=PA∩BPB=1615=56PB/A=PA∩BPA=1613=36=12

Q9.

Answer :

Consider the given events.
A = Both the children are female.
B = The elder child is a female.
C = At least one child is a male.
D = Both children are male.

Clearly, S=M1M2, M1F2, F1M2, F1F2A=F1F2B=F1M2, F1F2C=M1F2, F1M2, M1M2 D=M1M2
[Here, first child is elder and second is younger]

D∩C=M1M2 and A∩B=F1F2i Required probability = PD/C=nD∩CnC=13ii Required probability = PA/B=nA∩BnB=12

Page 31.21 Ex. 31.2

Q1.

Answer :

Consider the given events.
A = A king in the first draw
B = A king in the second draw

Now, PA=452=113PB/A= Getting a king in the second draw after getting a king in the first draw =351 After the first draw, the total number of cards will be 51. Then, 3 kings will be remaining. =117∴ Required probability = PA∩B=PA×PB/A=113×117=1221

Q2.

Answer :

Consider the given events.
A = An ace in the first draw
B = An ace in the second draw
C = An ace in the third draw
D = An ace in the fourth draw

Now, PA=452=113PB/A=351=117PC/A∩B=250=125PD/A∩B∩C=149∴ Required probability = PA∩B∩C∩D=PA×PB/A×PC/A∩B×PD/A∩B∩C =113×117×125×149

Q3.

Answer :

Consider the given events.
A = A white ball in the first draw
B = A white ball in the second draw

Now, PA=712PB/A=611∴ Required probability = PA∩B=PA×PB/A=712×611=722

Q4.

Answer :

There are 12 even numbers between 1 to 25.

Consider the given events.
A = An even number ticket in the first draw
B = An even number ticket in the second draw

Now, PA=1225PB/A=1124∴ Required probability = PA∩B=PA×PB/A=1225×1124=1150

Q5.

Answer :

Consider the events
A = An ace in the first draw
B = An ace in the second draw
C = Getting an ace in the third draw

Now, PA=1352=14PB/A=1251=417PC/A∩B=1150∴ Required probability = PA∩B∩C =PA×PB/A×PC/A∩B =14×417×1150 =11850

Q6.

Answer :

(i) Consider the given events
A = A king in the first draw
B = A king in the second draw

Now, PA=452=113PB/A=351=117∴ Required probability = PA∩B =PA×PB/A =113×117 =1221

(ii) Consider the given events
A = A king in the first draw
B = An ace in the second draw

Now, PA=452=113PB/A=451=451∴ Required probability = PA∩B =PA×PB/A =113×1451 =4663

(iii) Consider the given events.
A = A heart in the first throw
B = A red card in the second throw

Now,PA=1352=14PB/A=2551∴ Required probability = PA∩B =PA×PB/A =14×2551 =25204

Q7.

Answer :

There are 10 even numbers and 10 odd numbers between 1 to 20.

Consider the given events.
A = An even number in the first draw
B = An odd number in the second draw

Now, PA=1020=12PB/A=1019∴ Required probability=PA∩B=PA×PB/A=12×1019=519

Q8.

Answer :

(i) Consider the given events.
A = A white or red ball in the first draw
B = A white or red ball in the second draw

Now, PA=712PB/A=611∴ PA∩B=PA×PB/A =712×611 =722∴ Required probability = 1-PA∩B =1-722 =1522

(ii) Consider the given events.
A = A white ball in the first draw
B = A black ball in the second draw
C = A red ball in the third draw

Now, PA=416=14PB/A=715PC/A∩B=514∴ Required probability = PA∩B∩C=PA×PB/A×PC/A∩B =14×715×514 =124

Q9.

Answer :

Consider the given events.
A = A white or black ball in the first draw
B = A white or black ball in the second draw
C = A white or black ball in the third draw

Now, PA=815PB/A=714=12PC/A∩B=613∴ Required probability = PA∩B∩C =PA×PB/A×PC/A∩B =815×12×613 =865

Q10.

Answer :

Consider the given events.
A = A heart in the first draw
B = A diamond in the second draw

Now, PA=1352=14PB/A=1351∴ Required probability = PA∩B=PA×PB/A=14×1351=13204

Q11.

Answer :

Consider the given events.
A = A black ball in the first draw
B = A black ball in the second draw

Now, PA=1015=23PB/A=914∴ Required probability = PA∩B=PA×PB/A=23×914=37

Q12.

Answer :

Consider the given events.
A = A king in the first draw
B = A king in the second draw
C = An ace in the third draw
Now,PA=452=113PB/A=351=117PC/A∩B=450=225∴ Required probability = PA∩B∩C =PA×PB/A×PC/A∩B =113×117×225 =25525

Q13.

Answer :

Consider the given events.
A = A good orange in the first draw
B = A good orange in the second draw
C = A good orange in the third draw
Now, PA=1215=45PB/A=1114PC/A∩B=1013∴ Required probability = PA∩B∩C∩D =PA×PB/A×PC/A∩B =45×1114×1013 =4491

Page 31.32 Ex. 31.3

Q1.

Answer :

Given:PA=713PB=913 PA∩B=413Now,PAB=PA∩BPB⇒ PAB=413913=49

Q2.

Answer :

Given:PA=0.6PB=0.3 PA∩B=0.2Now,PAB=PA∩BPB⇒ PAB=0.20.3=23PBA=PA∩BPA⇒ PBA=0.20.6=13

Q3.

Answer :

Given:PB=0.5 PA∩B=0.32Now,PAB=PA∩BPB⇒ PAB=0.320.5=3250=0.64

Q4.

Answer :

Given:PA=0.4PB=0.8 PB/A=0.6Now,PB/A=PA∩BPA⇒0.6=PA∩B0.4⇒PA∩B=0.24PA/B=PA∩BPB=0.240.8=0.3PA∪B=PA+PB-PA∩B⇒PA∪B=0.4+0.8-0.24=0.96

Q5.

Answer :

i) Given: PA=13PB=14 PA∪B=512PA∪B=PA+PB-PA∩B⇒512=13+14-PA∩B⇒PA∩B=13+14-512=212=16Now,PA/B=PA∩BPB=1614=46=23PB/A=PA∩BPA=1613=36=12

ii) Given: PA=611PB=511PA∪B=711PA∪B=PA+PB-PA∩B⇒711=611+511-PA∩B⇒PA∩B=611+511-711=411Now,PA/B=PA∩BPB=411511=45PB/A=PA∩BPA=411611=46=23

iii) Given: PA=713PB=913 PA∩B=413Now,PA/B=PA∩BPB=413913=49PA/B=1- PA/B=1-49=59

Q6.

Answer :

Given: 2PA=PB=513 PA/B=25∴ PA=526 PB=513Now, PA/B=PA∩BPB⇒25=PA∩B513⇒PA∩B=25×513=213PA∪B=PA+PB-PA∩B =526+513-213 =1126

Q7.

Answer :

Given: PA=611PB=511 PA∪B=711(i) PA∪B=PA+PB-PA∩B⇒711=611+511-PA∩B⇒PA∩B=611+511-711=411ii PA/B=PA∩BPB =411511 =45iii PB/A=PA∩BPA =411611 =46 =23

Q8.

Answer :

(i) Consider the given events.
A = Heads on third toss
B = Heads on first two tosses

Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (T, T, H)}
B = {(H, H, H), (H, H, T)}

Now, A∩B=H, H, H∴ Required probability=PA/B=nA∩BnB=12

(ii) Consider the given events.
A = At least two heads
B = At most two heads

Clearly,
A = {(H, H, H), (H, T, H), (T, H, H), (H, H, T)}
B = {(T, T, T), (H, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T)}

Now, A∩B=H, T, H, T, H, H, H, H, T∴ Required probability = PA/B=nA∩BnB=37

(iii) Consider the given events.
A = At most two tails
B = At least one tail

Clearly,
A = {(T, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T), (H, H, H)}
B = {(T, T, T), (T, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T)}

Now, A∩B={(T, T, H), (T, H, H), (H, H, T), (T, H, T), (H, H, T), (H, T, T)}

∴ Required probability=PA/B=nA∩BnB=67

Q9.

Answer :

(i) Consider the given events.
A = Tail appears on one coin
B = One coin shows head

Clearly,
A = {(H, T), (T, H)}
B = {(H, T), (T, H)}

Now, A∩B=H, T, T, H∴ Required probability=PA/B=nA∩BnB=22=1

(ii) Consider the given events.
A = No tail appears
B = No head appears

Clearly,
A = {(H, H)}
B = {(T, T)}

Now, A∩B=ϕ∴ Required probability = PA/B=nA∩BnB=01=0

Q10.

Answer :

Consider the given events.
A = Getting 4 on third throw
B = Getting 6 on first throw and and 5 on second throw

Clearly,
A = {(1, 1, 4), (1, 2, 4), (1, 3, 4), (1, 4, 4), (1, 5, 4), (1, 6, 4), (2, 1, 4), (2, 2, 4), (2, 3, 4), (2, 4, 4), (2, 5, 4), (2, 6, 4),
. .
. .
. .
(6, 1, 4), (6, 2, 4), (6, 3, 4), (6, 4, 4), (6, 5, 4), (6, 6, 4)}
B = {6, 5, 1), (6, 5, 2), (6, 5, 3), (6, 5, 4), (6, 5, 5), (6, 5, 6)}

Now, A∩B=6, 5, 4∴ Required probability = PA/B=nA∩BnB=16∴ Required probability = PB/A=nA∩BnA=136

Q11.

Answer :

Consider the given events.
A = Son standing on one end
B = Father standing in the middle

Clearly, S=MFS, MSF, FSM, FMS, SMF, SFMA=MFS, FMS, SMF, SFM, B=MFS, SFM

Now,A∩B=MFS, SFM i Required probability = PA/B=nA∩BnB=22=1ii Required probability = PB/A=nA∩BnA=24=12

Q12.

Answer :

Consider the given events.
A = 4 appears on the die
B = The sum of the numbers on two dice is 6.

Clearly,
A = {(1, 4) (2, 4), (3, 4),(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 4), (6, 4)}
B = {(1, 5), (5, 1), (2, 4), (4, 2),(3, 3)}

Now, A∩B=2,4 and 4,2∴ Required probability = PA/B=nA∩BnB=25

Q13.

Answer :

Consider the given events.
A = 4 appears on second die
B = The sum of the numbers on two dice is 8.

Clearly,
A = {(1, 4), (2, 4), (3, 4), (4, 4) (5, 4) (6, 4)}
B = {(4, 4), (3, 5), (5, 3) (2, 6), (6, 2)}

Now, A∩B=4,4 ∴ Required probability = PB/A=nA∩BnA=16

Q14.

Answer :

Consider the given events.
A = Number appearing on second die is odd
B = The sum of the numbers on two dice is 7.

Clearly,
A = {(1, 1), (1, 3),(1, 5), (2, 1), (2, 3), (2, 5), (3, 1), (3, 3), (3, 5), (4, 1), (4, 3), (4, 5), (5, 1), (5, 3), (5, 5),(6, 1), (6, 3), (6, 5)}
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}

Now, A∩B=2,5,4,3,6,1∴ Required probability = PB/A=nA∩BnA=318=16

Q15.

Answer :

Consider the given events
A = A prime number appears on second die.
B = The sum of the numbers on two dice is 7.

Clearly,
A = {(1, 2), (1, 3), (1, 5), (2, 2), (2, 3)(2, 5), (3, 2), (3, 3), (3, 5) (4, 2), (4, 3), (4, 5),(5, 2), (5, 3), (5, 5), (6, 2), (6, 3),(6, 5)}
B = {(2, 5), (5, 2), (3, 4), (4, 3), (1, 6), (6, 1)}

Now, A∩B={(2, 5), (5, 2),(4, 3)}

∴ Required probability = PB/A=nA∩BnA=318=16

Q16.

Answer :

Consider the given events.
A = The number is odd
B = The number is prime

Clearly,
A = {1, 3, 5}
B = {2, 3,5}

Now, A∩B=3,5∴ Required probability = PB/A=nA∩BnA=23

Q17.

Answer :

Consider the given events.
A = 4 appears on first die
B = The sum of the numbers on two dice is 8 or more.

Clearly,
A = {(4, 1), (4, 2), (4, 3), (4, 4) (4, 5), (4, 6)}
n(A) = 6

B = {(2, 6), (3, 5), (3, 6), (4, 4), (4, 5), (4, 6),(5, 3), (5, 4), (5, 5) (5, 6), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n(B) = 15

Now, A∩B=4, 4, 4, 5, 4, 6∴ Required probability = PB/A=nA∩BnA=36=12

Q18.

Answer :

Consider the given events.
A = At least one die does not show 5
B = The sum of the numbers on two dice is 8.

Clearly,
A = {(1, 1), (1, 2) (1, 3), (1, 4),(1, 6),(2, 1), (2, 2) (2, 3), (2, 4), (2, 6), (3, 1), (3, 2), (3, 3) (3, 4), (3, 6),(4, 1), (4, 2), (4, 3), (4, 4), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 6)}
B = {(2, 6), (3, 5), (4, 4), (5, 3),(6, 2)}

Now, A∩B=4, 4, 6, 2, 2, 6∴ Required probability=PB/A=nA∩BnA=325

Q19.

Answer :

Suppose O represents the event of getting two odd numbers and S represents the event of getting their sum as an even number.Now,PO/S=PO∩S PS=5C29C24C2+5C29C2=5C24C2+5C2=1016=58

Q20.

Answer :

Consider the given events.
A = 5 appears on the die at least once
B = The sum of the numbers on two dice is 8.

Clearly,
A = {(1, 5),(2, 5),(3, 5),(4, 5),(5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}
B = {(2, 6), (3, 5), (4, 4), (5, 3),(6, 2)}

Now, A∩B=3,5,5,3∴ Required probability = PA/B=nA∩BnB=25

Q21.

Answer :

Consider the given events.
A = First die shows 6
B = The sum of the numbers on two dice is 7.

Clearly,
A= {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(2, 5), (5, 2), (4, 3), (3, 4), (1, 6), (6, 1)}

Now, A∩B=6, 1∴ Required probability = PB/A=nA∩BnA=16

Q22.

Answer :

Consider the given events.
E = The sum of the numbers on two dice is 10 or more
F = 5 appears on first die

Clearly,
E = {(4, 6),(5, 5),(5, 6),(6, 4), (6, 5), (6, 6)}
F = {(5, 1), (5, 2), (5, 3), (5, 4) (5, 5), (5, 6)}

Now, E∩F=5, 5, 5, 6∴ Required probability = PE/F=nE∩FnF=26=13

Second case:
Consider the given events.
E = The sum of the numbers on two dice is 10 or more
F = 5 appears on a die at least once

Clearly,
E = {(4, 6),(5, 5),(5, 6),(6, 4), (6, 5), (6, 6)}
F = {(1, 5),(2, 5),(3, 5),(4, 5),(5, 5), (6, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6)}

Now, E∩F=5, 5, 5, 6, 6, 5∴ Required probability = PE/F=nE∩FnF=311

Page 31.33 Ex. 31.3

Q23.

Answer :

Consider the given events.
M = Students passes Mathematics
C = Students passes Computer Science

We have, PM=45 PM∩C=12Now,PCM=PM∩CPM =1245=58

Q24.

Answer :

Suppose S represents the event of buying a shirt and T represents the event of buying a trouser.We have, PS=0.2PT=0.3 PS/T=0.4Now,PS/T=PS∩T PT⇒PS∩T= PS/T× PT=0.4×0.3=0.12PT/S=PS∩T PS=0.120.2=0.6

Q25.

Answer :

Suppose S represents a student chosen randomly studying in class XII and G represents a female student chosen randomly.We have, PG=4301000 PS/G=431000Now,PS/G=PS∩G PG=4310004301000=110

Q26.

Answer :

Sample space, S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Consider the given events.
A = Even number appears on the card
B = A number, which is more than 3, appears on the card

Here,
A = {2, 4, 6, 8, 10}
B = {4, 5, 6, 7, 8, 9, 10}

Now, A∩B=4,6,8,10∴ Required probability=PA/B=nA∩BnB=47

Page 31.49 Ex. 31.4

Q1.

Answer :

S=H H H H H T H T H H T T T H H T H T T T H T T Ti PA=48=12PB=48=12Now, PA∩B=28=14 PA∩B=PAPBThus, A and B are independent events.ii PA=48=12PB=48=12Now, PA∩B=08=0 PA∩B ≠ PAPBThus, A and B are not independent events.iii PA=38PB=48=12Now, PA∩B=28=14 PA∩B ≠ PAPBThus, A and B are not independent events.

Q2.

Answer :

Total number of events=36P4 on first die=PA=636=16P5 on second die=PB=636=16PA∩B=136PA∩B=PAPBThus, A and B are independent events.

Q3.

Answer :

i) Pking or queen=PA=852=213Pqueen or jack= PB=852=213PA∩B=Pqueen=452=113PA∩B ≠ PA PBThus, A and B are not independent events.
ii) Pblack=PA=2652=12Pking= PB=452=113PA∩B=Pblack king=252=126PA∩B = PA PBThus, A and B are independent events.
iii) Pspade=PA=1352=14Pace= PB=452=113PA∩B=Pace of spade=152PA∩B =PA PBThus, A and B are independent events.

Q4.

Answer :

S=H H H H H T H T H H T T T H H T H T T T H T T Ti PA=48=12PB=48=12PA∩B=28=14=PAPBThus, A and B are independent events.ii PC=28=14PB=48=12PB∩C=28=14≠ PBPCThus, B and C are not independent events.iii PC=14PA=12 PC∩A=18=PCPAThus, A and C are independent events.

Q5.

Answer :

PA∪B =PA+ PB-PA∩B⇒PA∩B =PA+ PB-PA∪B⇒PA∩B =14+13-12⇒PA∩B =3+4-612⇒PA∩B =112=14×13=PAPBThus, A and B are independent events.

Q6.

Answer :

Given: A and B are independent events.PA=0.3PB=0.6i PA∩B =PA PB=0.3×0.6=0.18ii PA∩B¯ =PA PB¯=PA1-PB=0.3×1-0.6=0.3×0.4=0.12iii PA¯∩B =PA¯ PB=PB1-PA=0.6×0.7=0.42iv PA¯∩B¯ =PA¯ PB¯=1-PA1-PB=0.7×0.4=0.28v PA∪B =PA+ PB-PA∩B=0.3+0.6-0.18=0.9-0.18=0.72vi PAB=PA∩BPB=0.180.6=0.3vii PBA=PA∩BPA=0.180.3=0.6

Q7.

Answer :

PB¯=0.65⇒1-PB=0.65⇒PB=1-0.65=0.35Now,PA∪B=PA+PB-PA∩B⇒PA∪B=PA+PB-PA×PB⇒0.85=PA+0.35-0.35×PA⇒0.85-0.35=PA1-0.35⇒PA=0.50.65=0.77

Q8.

Answer :

Let:PA=x PB=yPA¯ ∩B=215⇒PA¯ ×P B=215⇒1-xy=215 …1PA ∩B¯=16⇒PA ×P B=16 ⇒1-yx=16 …2Subtracting 2 from 1, we getx-y=130x=y+130Substituting the value of x in 2, we gety+1301-y=16⇒30y2-29y+4=0⇒y=16, 45

Q9.

Answer :

PA∩B=PA PB A and B are independent events16=PA PB⇒PA=16PB …1PA¯∩B¯=1-PA1-PB⇒13=1-PA1-PB⇒13=1-16PB1-PB Using 1Let PB=x⇒6x-16×1-x=13⇒6x-1-6×2+x=2x⇒6×2-5x+1=0⇒2x-13x-1=0⇒x=12 or x=13If PB=12, then PA=13If PB=13, then PA=12

Q10.

Answer :
PA∪B =P A + PB-PA∩BPA∪B =P A + PB-P A×PB ∵ A and B are independent events⇒0.6=0.2 +PB-0.2×PB⇒0.6-0.2=PB1-0.2⇒PB=0.6-0.21-0.2⇒PB=0.40.8⇒PB=12⇒PB=0.5

Q11.

Answer :

S= {1 1, 1 2, 1 3, 1 4, 1 5, 1 6, 2 1, 2 2, 2 3, 2 4, 2 5, 2 6, 3 1, 3 2, 3 3, 3 4, 3 5, 3 6, 4 1, 4 2, 4 3, 4 4, 4 5, 4 6, 5 1, 5 2, 5 3, 5 4, 5 5, 5 6, 6 1, 6 2, 6 3, 6 4, 6 5, 6 6}nS= 36E=Getting a number greater than 3 on each toss =4 4, 4 5, 4 6, 5 4, 5 5, 5 6, 6 4, 6 5, 6 6nE=9PE=936 =14

Q12.

Answer :

PA solving the problem=PA=23PB solving the problem=PB=35We need to find out if PA¯ ∩B¯=PA¯ PB¯ A and B are independent events =1-PA1-PB=1-231-35=13×25=215

Q13.

Answer :

PA=P4, 5 or 6 on first toss=1836=12PB=P1, 2, 3 or 4 on second toss=2436=23It is clear that A and B are independent events.⇒PA∩B=PAPB⇒PA∩B=12×23∴ PA∩B=13

Page 31.50 Ex. 31.4

Q14.

Answer :

Given: Bag contains 3 red and 2 black balls.Let three red balls be R1, R2 and R3 and 2 black balls be B1 and B2.Sample space: {R1,R2, R1,R3, R1, B1, R1, B2, R1, B3R2, R3, R2, B1, R2, B2R3, R3iPdrawing two red balls =925iiPdrawing two black = 425iiiPfirst red and second black = 625

Q15.

Answer :

Pking=452Pqueen=452Pjack=452These cards can be drawn in 3P3 ways.Pking, queen and jack=452×452×452×3P3=3!2197=62197

Q16.

Answer :

Let:A=Particle X is defectiveB=Particle Y is defective∴ P(A)=9100 P(B)=5100Required probability=PA¯∩B¯ =PA¯×PB¯ =1-PA×1-PB =1-9100×1-5100 =91100×95100 =0.91×0.95 =0.8645

Q17.

Answer :

PA=PA hits target=13PB=PB hits target=25Now,PA∪B=Ptarget will be hit by either A or B⇒PA∪B=PA+PB-PA∩B ⇒PA∪B=PA+PB-PAPB A and B are independent⇒PA∪B=13+25-13×25⇒PA∪B=5+6-215⇒PA∪B=915⇒PA∪B=35

Q18.

Answer :

P gun hits the plane=1-gun does not hit the plane⇒PA=1-PA¯Now,⇒PA¯=1-0.41-0.31-0.21-0.1 =0.6×0.7×0.8×0.9 =0.3024∴ PA=1-0.3024 =0.6976

Q19.

Answer :

The odds against event A are 5 to 2.PA = 25+2=27The odds in favour of event B are 6 to 5.PB=66+5=611i Patleast one event occurs =PA∪B=PA+PB-PA∩B=PA+PB-PA×PB=27+611-27×611=22+4277-1277=22+42-1277=5277∴ PA∪B=5277ii Pnone of the event occurs=1-PA∪B=1-5277=2577

Q20.

Answer :

Pgetting an odd number in one throw=12Here, getting an odd number in three throws refers to 3 independent events.PA=PB=PC=12PA∪B∪C=PA+PB+PC-PA∩B+B∩C+C∩A+PA∩B∩C∩=12+12+12-12×12+12×12+12×12+12×12×12=32-34+18=12-6+18=78

Q21.

Answer :

Total balls =10 black + 8 red balls = 18 ballsPfirst red ball = 818Psecond red ball = 818Pfirst ball is black=1018Psecond ball is black=1018i Ptwo red balls= 818×818=1681ii Pfirst ball is black and second is red= 1018×818=2081iii Pone of them is black and other is red= Pfirst ball is red and second is black+Pfirst ball is black and second is red=818×1018+1018×818=2081+2081=4081

Q22.

Answer :

Total balls=4 red balls +7 blue balls=11 ballsi P2 red balls= Pfirst ball is red×Psecond ball is red=411×411=16121ii P2 blue balls=Pfirst ball is blue×Psecond ball is blue=711×711=49121iii Pone red and one blue=Pfirst red and second blue+Pfirst blue and second red=411×711+711×411=28121+28121=56121Disclaimer: In the question, instead of black balls it should be blue balls.

Q23.

Answer :

PA coming in time = 37PA not coming in time = 1-37=47PB coming in time = 57PB not coming in time = 1-57=27Ponly one of A and B coming in time = PA PB¯ + PA¯PB=37×27+47×57=649+2049=2649

Page 31.63 Ex. 31.5

Q1.

Answer :

Given:Bag 1=3W+6B ballsBag 2=5B+4W ballsPballs of same colour are drawn=Pboth black+Pboth white=69×59+39×49=3081+1281=4281=1427

Q2.

Answer :

Given:Bag 1=3R+5B ballsBag 2=6R+4B ballsPone is red and one is black=Pred from bag 1 and black from bag 2+Pred from bag 2 and black from bag 1=38×410+58×610=1280+3080=4280=2140

Q3.

Answer :

Given: Box=10B+8R ballsi Pboth red balls=818×818=64324=1681ii Pfirst black and second red=818×1018=80 324=2081iii Pone is red and one is black=Pfirst red and second black +Pfirst red and second black=8’18×1018+1018×818=80324+80324=4081

Q4.

Answer :

Pexactly one ace=Pfirst card is ace +Psecond card is ace=452×4851+4852×451=192+19252×51=38452×51=32221

Q5.

Answer :

PBoth narrating different incident=PB lies and A speaks the truth+PA lies and B speaks the truth=PA∩B¯+PA¯∩B=PAPB¯+PA¯PB=0.751-0.8 +1-0.750.8=0.75×0.2+0.25×0.8=0.15+0.2=0.35=35%

Q6.

Answer :

PKamal gets selected= PA=13PMonica gets selected= PB=15i Pboth get selected =PA×PB=13×15=115ii Pnone of them get selected=PA¯×PB¯=1-PA1-PB=1-131-15=23×45=815iii Patleast one of them gets selected = PA∪B=PA+PB-PA∩B=PA+PB-PA×PB=13+15-13×15=13+15-115=715iv Pone of them gets selected=PA¯PB+PB¯PA=PB1-PA+PA1-PB=151-13+131-15=215+415=615=25

Q7.

Answer :

Given: Box= 3 W+4R+5B ballsPone white and one black=Pfirst white and second black+Pfirst black and second white=312×511+512×311=15132+15132=30132=522

Q8.

Answer :

Patleast 2 balls are green=1-Pat most one ball is green=1-Pfirst green+Psecond green+Pthird green+Pno green=1-614×813×712+814×613×712+814×713×612+814×713×612=1-3362184+3362184+3362184+3362184=1-13442184=8402184=513

Q9.

Answer :

PArun gets selected=PA=14PTarun gets rejected=PB¯=23⇒PTarun gets selected=1-23=13Patleast one of them is selected=PA∪B⇒PA∪B=PA+PB-PA∩B=PA+PB-PA×PB=14+13-14×13=3+4-112=12

Q10.

Answer :

PB winning the game=Phead at 2nd turn+Phead at 4th turn+ …=12×12+12×12×12×12+ …=122+124+126+128+ …=141+122+124+126+ … =1411-14 For infinite GP: 1+a+a2+a3+ … =11-a=14×43=13

Q11.

Answer :

Pone red and one black=Pfirst red and second black+Pfirst black and second red=2652×2651+2652×2651 Without replacement=1351+1351=2651

Q12.

Answer :

We know that 5 and 10 are multiples of 5, while 4 and 8 are multiples of 4.Pmultiple of 5=210=15Pmultiple of 4=210=15Pmultiple of 5 and multiple of 4=Pmultiple of 5 on first card and multiple of 4 on second card +Pmultiple of 4 on first card and multiple of 5 on second card=210×29+210×29 Without replacement=490+490=890=445

Q13.

Answer :

It is given that the husband lies in 30% of the cases, while the wife lies in 35% cases

Pboth will contradict each other on the same fact=Phusband lies but wife tells the truth+Pwife lies but husband tells the truth=0.3×1-0.35+1-0.3×0.35=0.3×0.65+0.7×0.35=0.195+0.245=0.44=44%

Q14.

Answer :

Phusband will be selected=PA=17Pwife will be selected=PB=15i Pboth will be selected=PA∩B=PA×PB=17×15=135ii Ponly one of them will be selected=PAPB¯+PA¯PB=171-15+151-17=435+635=1035=27iii Pnone of them will be selected=PA∩B=PA¯×PB¯=1-171-15=2435

Q15.

Answer :

Given: Bag=7W+5B+4R ballsPatleast 3 balls are black=Pexactly 3 black+Pall 4 black=1116×515×414×313×4+516×415×314×213=1114×13+12×14×13=22+1364=23364

Q16.

Answer :

PA speaks truth=34PB speaks truth=45PC speaks truth=56Pmajority speaks truth=Ptwo speak truth+Pall speak truth=PA×PB1-PC+PA×PC1-PB+PC×PB1-PA+PA×PB×PC=34×451-56+34×561-45+45×561-34+34×45×56=12120+15120+20120+60120=107120

Page 31.64 Ex. 31.5

Q17.

Answer :

Given:Bag 1=4W+2B ballsBag 2=3W+5B ballsi Pboth are white=46×38=14ii Pboth are black=26×58=524iii Pone is white and one is black=Pwhite from bag 1 and black from bag 2+Pwhite from bag 2 and black from bag 1=46×58+38×26=2048+648=2648=1324

Q18.

Answer :

Given: Bag=4W+5R+7B ballsPatleast 2 white balls=1-Pmaximum 1 white ball=1-Pno white+Pexactly one white=1-1216×1216×1216×1216+416×1216×1216×1216×4=1-81256+108256=1-189256=256-189256=67256

Q19.

Answer :

Pking=PA=452Pqueen=PB=452Pjack=PC=452Pking, queen and jack=3!×PA×PB×PC =3×2×452×452×452=62197

Q20.

Answer :

Given:Bag A=4R+5B ballsBag B=3R+7B ballsi Pballs of different colours=Pred from bag A and black from bag B+Pred from bag B and black from bag A=49×710+310×59=2890+1590=4390ii Pballs of same colour=Pboth red +Pboth black=49×310+710×59=1290+3590=4790

Q21.

Answer :

PA hits the target=36PB hits the target=26PC hits the target=44=1Patleast 2 shots hit\=Pexactly 2 shots hit+Pall 3 shots hit=361-26+261-36+36×26×1 Here, the probability of C hitting the target is 1. So, it will always hit.When exactly 2 shots are hit, then either A hits or B hits.=36×46+26×36+636=12+6+636=2436=23

Q22.

Answer :

PA passing examination=29PB passing examination=59i Ponly A passing examination=PA passes PB fails=291-59=29×49=881ii Ponly one of them passing examination=PA passes and B fails+PB passes and A fails=29×1-59+59×1-29=881+3581=4381

Q23.

Answer :

Given:Urn A 4R+3BUrn B 5R+4BUrn C 4R+4BPtwo red and one black=Pblack from urn A+Pblack from urn B+Pblack from urn C=37×59×48+47×49×48+47×59×48=542×16126×20126=15+16+20126=51126=1742

Q24.

Answer :

PA grade in Maths=PA=0.2PA grade in Physics=PB=0.3PA grade in Chemistry=PC=0.5i Pgrade A in all subjects=PA×PB×PC=0.2×0.3×0.5=0.03ii Pgrade A in no subject=PA¯×PB¯×PC¯=1-0.2×1-0.3×1-0.5=0.8×0.7×0.5=0.28iii Pgrade A in two subjects=Pnot grade A in Maths+Pnot grade A in Physics+Pnot grade A in Chemistry=PA×PB×PC+PA×PB×PC+PA×PB×PC=1-0.2×0.3×0.5+0.2×1-0.3×0.5+0.2×0.3×1-0.5=0.8×0.3×0.5+0.2×0.7×0.5+0.2×0.3×0.5=0.12+0.07+0.03=0.22

Q25.

Answer :

Total number of events =36Pgetting 9=436=19PA winning=Pgetting 9 in first throw+Pgetting 9 in third throw+ …=19+1-191-19×19+ …=191+6481+64812+ … =1911-6481 1+a+a2+a3+ … =11-a=19×8117=917PB winning=Pgetting 9 in second throw+Pgetting 9 in fourth throw+ …=1-1919+1-191-191-19×19+ …=8811+6481+64812+ … =88111-6481 1+a+a2+a3+ … =11-a=881×8117=817∴ Winning ratio of A to B=917817=98

Q26.

Answer :

PA winning=Phead in first toss+Phead in fourth toss+ …=12+12×12×12×12+ …=121+123+126+ … =1211-18 1+a+a2+a3+ … =1 1-a=12×87=47PB winning=Phead in second toss+Phead in fifth toss+ …=12×12+12×12×12×12×12+ …=141+123+126+ … =1411-18 1+a+a2+a3+ … =11-a=14×87=27PC winning=Phead in third toss+Phead in sixth toss+ …=12×12×12+12×12×12×12×12×12+ …=181+123+126+ … =1811-18 1+a+a2+a3+ … =11-a=18×87=17

Q27.

Answer :

Psix=16Pno six=56PA winning=P6 in first throw+P6 in fourth throw+ …=16+56×56×56×16+ …=161+563+566+ … =1611-125216 1+a+a2+a3+ … =11-a=16×21691=3691PB winning=P6 in second throw+P6 in fifth throw+ …=56×16+56×56×56×56×16+ …=5361+563+566+ …=53611-125216 1+a+a2+a3+ … =11-a=536×21691=3091PC winning=P6 in third throw+P6 in sixth throw+ …=56×56×16+56×56×56×56×56×12+ …=252161+563+566+ … =2521611-125216 1+a+a2+a3+ … =11-a=25216×21691=2591

Q28.

Answer :

There are only three possible cases, wherein the sum of the numbers obtained after throwing 2 dice is 10, i.e. 4, 6 5, 5 6, 4.∴ Psum of the numbers is 10=336=112Psum of the numbers is not 10=1-112=1112Pany numner other than six=56PA winning=P10 in first throw+P10 in third throw+ …=112+1112×1112×112+ …=1121+11122+11124+ … =11211-121144 1+a+a2+a3+ … =11-a=112×14423=1223PB winning=1-PA winning=1123Now,PA winningPB winning=12231123=1211Hence proved.

Q29.

Answer :

It is given that bag A contains 3 red and 5 black balls 3R, 5B and bag B contains 2 red and 3 black balls 2R, 3B.Now,Pone red and 2 black= Pone red from bag A and two black from bag B+Pblack ball from bag A and remaining balls from bag B=38×35×24+58×25×34×2=980+3080=3980Note: 2 is multiplied by second term because there are two ways to select red and black balls from bag B. While the first way is to pick black ball first, followed by red, the second way is to pick red ball first, followed by black.

Q30.

Answer :

PFatima gets selected=PA=17PJohn gets selected=PB=15i Pboth of them get selected=PA∩B=PA×PB=17×15=135ii Ponly one of them gets selected=PA×PB¯+PA¯×PB=171-15+1-1715=17×45+67×15=435+635=1035=27iii Pnone of them get selected=PB¯×PA¯=1-15×1-17=45×67=2435

Q31.

Answer :

It is given that the bag contains 3 blue and 5 red marbles.i Pblue followed by red=38×58=1564ii Pred and blue in any order=Pblue followed by red+Pred followed by blue=38×58+58×38=1564+1564=3064=1532iii Psame colour=Pboth red+Pboth blue=58×58+38×38=2564+964=3464=1732

Q32.

Answer :

It is given that the urn contains 7 red and 4 black balls.i P2 red balls=711×711=49121ii P2 blue balls=411×411=16121iii Pone red ball and one blue ball=Pblue ball followed by red ball+Pred ball followed by blue ball=411×711+711×411=28121+28121=56121

Q33.

Answer :

i Pboth the cards are of same suit=Pboth the cards are of diamond+Pboth the cards are of spade+Pboth the cards are of club+Pboth the cards are of heart=1352×1352+1352×1352+1352×1352+1352×1352=116+116+116+116=416=14ii Pfirst ace and second red queen=Pace card×Pred queen=452×252=1338

Page 31.65 Ex. 31.5

Q34.

Answer :

i Pboth enter same section=Pboth enter section A+Pboth enter section B=40100×40100+60100×60100=425+925=1325ii Pboth enter different sections=1-Pboth enter same section=1-1325=1225

Q35.

Answer :

Pa six=16Pnot a six=1-16=56PA wins=P6 in first throw+P6 in third throw+ …=16+56×56×16+ …=161+562+564+ …=1611-2536 … 1+a+a2+a3+ … =11-a=16×3611=611PB wins=P6 in second throw+P6 in fourth throw+ …=56×16+56×56×56×16+ …=5361+562+564+ …=53611-2536 … 1+a+a2+a3+ … =11-a=536×3611=511

It can be seen that the probability that team A wins is not equal to the probability that team B wins.
Thus, the decision of the referee was not fair.

Page 31.72 Ex. 31.6

Q1.

Answer :

A black ball can be drawn in two mutually exclusive ways:

(I) By transferring a white ball from bag A to bag B, then drawing a black ball
(II) By transferring a black ball from bag A to bag B, then drawing a black ball

Let E1, E2 and A be the events as defined below:

E1 = A white ball is transferred from bag A to bag B
E2 = A black ball is transferred from bag A to bag B
A = A black ball is drawn

∴ PE1=511 PE2=611Now, PA/E1=38PA/E2=48Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =511×38+611×48 =1588+2488 =3988

Q2.

Answer :

A silver coin can be drawn in two mutually exclusive ways:

(I) Selecting purse I and then drawing a silver coin from it
(II) Selecting purse II and then drawing a silver coin from it

Let E1, E2 and A be the events as defined below:

E1 = Selecting purse I
E2 = Selecting purse II
A = Drawing a silver coin

It is given that one of the purses is selected randomly.

∴ PE1=12 PE2=12Now, PA/E1=26=13PA/E2=47Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =12×13+12×47 =16+27 =7+1242=1942

Q3.

Answer :

A yellow ball can be drawn in two mutually exclusive ways:

(I) By transferring a red ball from first to second bag, then drawing a yellow ball
(II) By transferring a yellow ball from first to second bag, then drawing a yellow ball

Let E1, E2 and A be the events as defined below:

E1 = A red ball is transferred from first to second bag
E2 = A yellow ball is transferred from first to second bag
A = A yellow ball is drawn

∴ PE1=59 PE2=49Now, PA/E1=610PA/E2=710Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =59×610+49×710 =3090+2890 =5890=2945

Q4.

Answer :

A white ball can be drawn in two mutually exclusive ways:

(I) Selecting bag I and then drawing a white ball from it
(II) Selecting bag II and then drawing a white ball from it

Let E1, E2 and A be the events as defined below:
E1 = Selecting bag I
E2 = Selecting bag II
A = Drawing a white ball

It is given that one of the bags is selected randomly.

∴ PE1=12 PE2=12Now,PA/E1=35PA/E2=26=13Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =12×35+12×13 =310+16 =9+530=1430=715

Q5.

Answer :

A white ball and a red ball can be drawn in three mutually exclusive ways:

(I) Selecting bag I and then drawing a white and a red ball from it
(II) Selecting bag II and then drawing a white and a red ball from it
(II) Selecting bag III and then drawing a white and a red ball from it

Let E1, E2 and A be the events as defined below:
E1 = Selecting bag I
E2 = Selecting bag II
E3 = Selecting bag II
A = Drawing a white and a red ball

It is given that one of the bags is selected randomly.

∴ PE1=13 PE2=13 PE3=13Now, PA/E1=1C1×3C16C2=315PA/E2=2C1×1C14C2=26PA/E3=4C1×3C112C2=1266Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2+ PE3PA/E3 =13×315+13×26+13×1266 =115+19+233 =33+55+30495=118495

Q6.

Answer :

Let E1, E2 and A be the events as defined below:
E1 = The coin shows a head
E2 = The coin shows a head
A = The noted number is 7 or 8

∴ PE1=12 PE2=12Now, PA/E1=1136PA/E2=211Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =12×1136+12×211 =1172+111 =121+72792=193792

Q7.

Answer :

Let A, E1 and E2 denote the events that the item is defective, machine A is selected and machine B is selected, respectively.

∴ PE1=60 100 PE2=40100Now,PA/E1=2100PA/E2=1100Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =60100×2100+40100×1100 =12010000+4010000 =120+4010000=16010000=0.016

Q8.

Answer :

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from bag A to bag B, then drawing a white ball
(II) By transferring a white ball from bag A to bag B, then drawing a white ball

Let E1, E2 and A be events as defined below:
E1 = A black ball is transferred from bag A to bag B
E2 = A white ball is transferred from bag A to bag B
A = A white ball is drawn

∴ PE1=715 PE2=815Now,PA/E1=510=12PA/E2=610=35Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =715×12+815×35 =730+825 =35+48150=83150

Q9.

Answer :

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a white ball
(II) By transferring a white ball from first to second bag, then drawing a white ball

Let E1, E2 and A be the events as defined below:
E1 = A black ball is transferred from first to second bag
E2 = A white ball is transferred from first to second bag
A = A white ball is drawn

∴ PE1=59 PE2=49Now,PA/E1=38PA/E2=48=12Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =59×38+49×12 =1572+29 =15+1672=3172

Page 31.73 Ex. 31.6

Q10.

Answer :

A white ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a white ball
(II) By transferring a white ball from first to second bag, then drawing a white ball

Let E1, E2 and A be the events as defined below:
E1 = A black ball is transferred from first to second bag
E2 = A white ball is transferred from first to second bag
A = A white ball is drawn

∴ PE1=59 PE2=49Now, PA/E1=614PA/E2=714Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =59×614+49×714 =30126+28126 =58126=2963

Q11.

Answer :

A white ball can be drawn in three mutually exclusive ways:

(I) By transferring two black balls from first to second urn, then drawing a white ball
(II) By transferring two white balls from first to second urn, then drawing a white ball
(III) By transferring a white and a black ball from first to second urn, then drawing a white ball

Let E1, E2, E3 and A be the events as defined below:
E1 = Two black balls are transferred from first to second bag
E2 = Two white balls are transferred from first to second bag
E2 = A white and a black ball is transferred from first to second bag
A = A white ball is drawn

∴ PE1=3C213C2=378 PE2=10C213C2=4578 PE3=10C1×3C113C2=3078Now, PA/E1=310PA/E2=510PA/E3=410Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2+PE3PA/E3 =378×310+4578×510+3078×410 =9780+225780+120780 =354780=59130

Q12.

Answer :

A red ball can be drawn in two mutually exclusive ways:

(I) By transferring a black ball from first to second bag, then drawing a red ball
(II) By transferring a red ball from first to second bag, then drawing a red ball

Let E1, E2 and A be the events as defined below:
E1 = A black ball is transferred from first to second bag
E2 = A red ball is transferred from first to second bag
A = A red ball is drawn

∴ PE1=814 PE2=614Now,PA/E1=815PA/E2=915Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =814×815+614×915 =64210+54210 =118210=59105

Page 31.84 Ex. 31.7

Q1.

Answer :

Let E1, E2 and E3 denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white and red.

∴ PE1=13 PE2=13 PE3=13Now, PA/E1=1C1×3C16C2=315=15PA/E2=2C1×1C14C2=26=13PA/E3=4C1×3C112C2=1266=211Using Bayes’ theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE3PA/E3 =13×1513×15+13×13+13×211 =1515+13+211=1533+55+30165=33118Required probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2++ PE3PA/E3 =13×1313×15+13×13+13×211 =1315+13+211=1333+55+30165=55118Required probability = PE3/A=PE3PA/E3PE1PA/E1+ PE2PA/E2++ PE3PA/E3 =13×21113×15+13×13+13×211 =21115+13+211=21133+55+30165=30118

Q2.

Answer :

Let A, E1 and E2 denote the events that the ball is red, bag A is chosen and bag B is chosen, respectively.

∴ PE1=12 PE2=12Now, PA/E1=35PA/E2=59Using Bayes’ theorem, we getRequired probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2 =12×5912×35+12×59 =2552

Q3.

Answer :

Let E₁, E₂ and E₃ denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the ball drawn is white.

∴ PE1=13 PE2=13 PE3=13Now, PA/E1=25PA/E2=35PA/E3=45Using Bayes’ theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE3PA/E3 =13×2513×25+13×35+13×45 =22+3+4=29

Q4.

Answer :

Let E₁, E₂ and E₃ denote the events of selecting Urn I, Urn II and Urn III, respectively.

Let A be the event that the two balls drawn are white.

∴ PE1=20100 PE2=60100 PE3=20100Now,PA/E1=7C210C2=2145PA/E2=4C210C2=645PA/E3=2C210C2=145Using Bayes’ theorem, we getRequired probability = PE3/A=PE3PA/E3PE1PA/E1+ PE2PA/E2++ PE3PA/E3 =20100×14520100×2145+60100×645+20100×145 =121+18+1=140

Q5.

Answer :

Let E₁, E₂ and E₃ denote the events that the doctor diagnoses correctly, the selected person suffers from TB and the selected person does not suffer from TB, respectively.

∴ PE1=11000 PE2=9991000Now, PA/E1=0.99PA/E2=0.001Using Bayes’ theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2 =11000×0.9911000×0.99+9991000×0.001 =110221

Q6.

Answer :

Let E₁ and E₂ denote the events that the first group and the second group win the competition, respectively. Let A be the event of introducing a new product.

P(E₁) = Probability that the first group wins the competition = 0.6

P(E₂) = Probability that the second group wins the competition = 0.4

P(A/E₁) = Probability of introducing a new product if the first group wins = 0.7

P(A/E₂) = Probability of introducing a new product if the second group wins = 0.3

The probability that the new product is introduced by the second group is given by P(E2/A).

Using Bayes’ theorem, we get
Required probability = PE2/A=PE2PA/E2PE1PA/E1+ PE2PA/E2 =0.4×0.30.6×0.7+0.4×0.3 =0.120.54=29

Q7.

Answer :

Let E₁, E₂ and E₃ denote the events that the person is a good orator, is a man and is a woman, respectively.

∴ PE1=12 PE2=12Now,PA/E1=5100PA/E2=251000Using Bayes’ theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2 =12×510012×5100+12×251000

Q8.

Answer :

Let E₁, E₂ and E₃ denote the events that the two consecutive letters are visible, the letter has come from LONDON and the letter has come from CLIFTON, respectively.

∴ PE1=12 PE2=12Now,PA/E1=25PA/E2=16Using Bayes’ theorem, we geti Required probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2 =12×2512×25+12×16 =2525+16=251730=1217ii Required probability = PE2/A=PE1PA/E1PE1PA/E1+ PE2PA/E2 =12×1612×25+12×16 =1625+16=161730=517

Q9.

Answer :

Let E₁, E₂ and E₃ denote the events that the IQ is more than 150, the selected student is a boy and the selected student is a girl, respectively.

∴ PE1=60100 PE2=40100Now, PA/E1=5100PA/E2=10100Using Bayes’ theorem, we getRequired probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2 =60100×510060100×5100+40100×10100

Q10.

Answer :

Let E₁, E₂ and E₃ denote the events that machine X produces bolts, machine Y produces bolts and machine Z produces bolts, respectively.

Let A be the event that the bolt is defective.

Total number of bolts = 1000 + 2000 + 3000 = 6000

P(E₁) = 10006000=16

P(E₂) =20006000=13

P(E₃) = 30006000=12

The probability that the defective bolt is produced by machine X is given by P (E₁/A).

Now, PA/E1=1%=1100PA/E2=1.5%=151000PA/E3=2%=2100Using Bayes’ theorem, we get Required probability = PE1/A=PE1PA/E1PE1PA/E1+ PE2PA/E2++ PE2PA/E2 =16×110016×1100+13×151000+12×2100 =1616+12+1=161+3+66=110

Page 31.90 (Very Short Answers)

Q1.

Answer :

To be divisible by 5 ones place sholud be 5There are 3 places remaining which can be filled in 3 !=6 waysSo, 6 numbers can be formed out of 1, 2, 3 and 5, which are divisible by 5.Total 4-digit numbers=4!=24P4-digit number divisible by 5=624=14

Q2.

Answer :

P4 or 5 on a die=26=13Pgetting 4 or 5 on each of the dice simultaneously=13×13×13=127

Q3.

Answer :

Total 3-digit numbers that can be made out of 0, 2, 4, 6 and 8=4×5×5 hundreds place cannot be filled with 0=100But 222, 444, 666 and 888 are four numbers, which have the same digits at all places.P3-digit number having same digits at all places=4100=125

Q4.

Answer :

A cube has total 6 faces.Total possible outcomes in 5 throws=6×6×6×6×6=65The only way of getting 7 is by getting two 2s and one 3.Total possible ways=P532!=5×4×3×2×12×1×2×1=30Now,Pgetting 7 in 5 throws=3065=564

Q5.

Answer :

Total possible outcomes=20C3Consecutive numbers chosen=(1, 2, 3), (2, 3, 4) … (18, 19, 20)So, there are 18 favouable cases.PA=18C203

Q6.

Answer :

Here, 6 boys and 6 girls can be arranged in a line in 12! ways.Total possible outcomes=12!Consider 6 girls as a single element X.Now, 6 boys and X can be arranged in a line in 7! ways and girls can be arranged in 6! ways among them.Pall girls are together=7!×6!12!=7×6×5×4×3×2×1×6×5×4×3×2×112×11×10×9×8×7×6×5×4×3×2×1=111×12=1132

Q7.

Answer :

A and B are two independent events.∴ PA∪B¯=PA+PB¯-PA∩B¯⇒0.8=0.3+1-PB-PA PB¯⇒0.5=1-PB-0.31-PB⇒0.5=1-PB-0.3+O.3PB⇒0.5=0.7-PB1-0.3⇒0.7PB=0.2⇒PB=0.20.7=27

Q8.

Answer :

Pface value is not more than 5 and not less than 2=46=23Pface value is not more than 5 and not less than 2 in 4 throws=23×23×23×23=1681

Page 31.91 (Very Short Answers)

Q9.

Answer :

Pexactly one of 2 events=PA∪B-PA∩B=PA+PB-PA∩B-PA∩B=PA+PB-2PA∩B

Q10.

Answer :

Number of cubes in first 100 natural numbers = 1,8,27,64So, there are 4 cubes in first 100 natural numbers.Pgetting a cube from a set of first 100 natural numbers=4100=125

Q11.

Answer :

PA wins=PAPB wins=PB=PA2PC wins=PC=PB2=PA4Now,PA+PB+PC=1⇒PA+PA2+PA4=1⇒PA4+2+14=1⇒PA=47PA loses=PA¯=1-47=37

Q12.

Answer :

A, B and C are mutually exclusive and exhaustive events.∴ PA+PB+PC=1

Q13.

Answer :

Disclaimer: the question seems to be incorrect.

Q14.

Answer :

A and B are two independent events.∴ PA∩B¯=PAPB¯=PA1-PB=PA-PAPB

Q15.

Answer :

PBA=PA∩BPA⇒0.5=PA∩B0.3⇒PA∩B=0.5×0.3⇒PA∩B=0.15Now,PA∪B=PA+PB-PA∩B =0.3+0.6-0.15 =0.9-0.15 =0.75

Page 31.91 (Multiple Choice Questions)

Q1.

Answer :

a 13 32Here, the three boxes contain 3 white and1black 3W, 1B, 2 white and 2 black 2W, 2B and 1 white and 3 blackballs1W, 3B, respectively.P2W, 1B=34×24×34+34×24×14+14×24×14=1864+664+264=2664=1332

Q2.

Answer :

a 4485×49Total cards=52There are four suits of cards in a pack, i.e. diamond, heart, spade and club.Pall 4 cards are of same suit=Pall 4 cards are of diamond+Pall 4 cards are of heart+Pall 4 cards are of spade+Pall 4 cards are of club=4×1352×1251×1150×1049=4×1185×49=4485×49

Q3.

Answer :

a 0.39PA∪B=PA+PB-PA∩B=0.25+0.5-0.14=0.61Pboth A and B not happening=PA∪B’=1-PA∪B=1-0.61=0.39

Q4.

Answer :

b 27100Pstudent gets first division=110Pstudent gets second division=35Pstudent gets third division=14Pstudents fails =Pstudent does not get first division×Pstudent does not get second division×Pstudent does not get third division =1-1101-351-14 =910×25×34 =54200 =27100

Q5.

Answer :

a 0.0875Here, there are total 5 ways by which India can get at least 7 points. 1 2 points+ 2 points + 2 points + 2 points=0.5×0.5×0.5×0.52 1 point+ 2 points + 2 points + 2 points = 0.05×0.5×0.5×0.53 2 points+ 1 point + 2 points + 2 points = 0.5×0.05×0.5×0.54 2 points+ 2 points + 1 point + 2 points= 0.5×0.5×0.05×0.55 2 points+ 2 points + 2 points + 1 point = 0.5×0.5×0.5×0.05Patleast 7 points=0.5×0.5×0.5×0.5 +40.05×0.5×0.5×0.5=0.0625 +40.00625=0.0625+0.025=0.0875

Q6.

Answer :

a 136Pyellow face=36=12Pred face=26=13Pone face=16Pyellow face, red face and blue face appear in the required order=12×13×16=136

Page 31.92 (Multiple Choice Questions)

Q7.

Answer :

b) 37

A leap year has 366 days

For a non-leap year:
52 weeks + 1 day

For a leap year:
52 weeks + 2 days

Sample space=Monday, Tuesday, Tuesday, Wednesday,Wednesday, Thursday,Thursday, Friday, Friday, Saturday, Saturday, Sunday, Sunday, MondayFavourable cases=3P53 Fridays or 53 Saturdays=37

Q8.

Answer :

d) 2324

4 letters can be placed in 4 envelopes in 4! ways = 24 ways
Now, there is only one method, by which all the letters are placed in the right envelope.

P(all letters are placed in the right envelopes) = 124

P(all letters are not placed in the right envelopes) = 1 – P(all letters are placed in the right envelopes)

=1-124=2324

Q9.

Answer :

a) 720PA speaks truth=0.75PA lies= 1-0.75=0.25PB speaks truth=0.8PB lies= 1-0.8=0.2Pcontradicting each other in a statement=P(A speaks truth and B lies)+PB speaks truth and A lies=0.75×0.2+0.8×0.25=0.15+0.2=0.35=35100=720

Q10.

Answer :

c) 1719Pproduct is even=1-Pproduct is oddFor the product to be odd, the two digits must be odd.Now, 10 numbers are odd in the first 20 integers.∴ Pproduct is even=1-1020×919×818=1-938×818=1-219=1719

Q11.

Answer :

(c) 1529

For sum of two integers to be odd, one integer should be even and the other should be odd.
In 30 consecutive integers, 15 are even and 15 are odd.

P(sum is odd) = P(first integer is odd and second is even) + P(first integer is even and second integer is odd)

=1530×1529+1530×1529=45030×29=1529

Q12.

Answer :

d) 23We know that the bag contains 5B (black), 4W (white) and 3R (red) balls.Now,PB=512PR=312PB or R=PB+PR =512+312 =812=23

Q13.

Answer :

a) 136Ppair of aces=Pace in first throw×Pace in second throw=16×16=136

Q14.

Answer :

a) 584Given:Red balls=2Blue balls=3Black balls=4Pall three balls are of same colour=P(all three are blue)+Pall three are black=39×28×17+49×38×27=184+484=584

Q15.

Answer :

b) dependentS=HHH, HHT, HTH, HTT, THH, THT, TTH, TTTPA=P2 heads=38PB=Plast one is head=48PA∩B=28=14≠PA PBThus, A and B are dependent.

Q16.

Answer :

a) P5775Total possible ways of leaving the lift =7×7×7×7×7=755 people can leave different floors in P57 ways.P5 people leaving the lift at different floors=P5775

Q17.

Answer :

a) 6464Pgood item=1016Pdefected item =616Peither good or defected item=Pgood item+Pdefected item=1016+616=1616= 1=6464

Q18.

Answer :

(c) 1116
Rusted items =3+5=8Rusted nails =3Total nails = 6Pgetting a rusted item or a nail=Pgetting a rusted item+Pgetting a nail-Pgetting a rusted item and a nail=816+616-316=8+6-316=1116

Page 31.93 (Multiple Choice Questions)

Q19.

Answer :

d) 48108Psame coloured socks=Pboth brown+Pboth white=59×48+49×38=2072+1272=3272=49=48108

Q20.

Answer :

a) 14PA=13PB⇒PB=3PA …1A and B are mutually exclusive events.⇒ PA∩B=0Now,PA∪B=PA+PB=PS⇒PA+PB=1⇒PA+3PA=1 From 1⇒4PA=1⇒PA=14

Q21.

Answer :

(d) P (B) − P (A ∩ B)

From the diagram, we get A∩B and A¯ ∩B are mutually exclusive events such that (A∩B) ∪(A¯∩B)=B. Therefore by addition theorem of probability we have P(A∩B) + P(A¯∩B) = P(B)∴ PA∩B=PB-PA∩B

Q22.

Answer :

d 0.9PA∪B=PA+PB-PA∩B⇒PA+PB=PA∪B+PA∩B⇒PA+PB=0.8 +0.3⇒PA+PB=1.1⇒1-PA¯+1-PB¯=1.1⇒PA+PB=2-1.1⇒PA+PB=0.9

Q23.

Answer :

(c) 8/15

A white ball can be drawn in two mutually exclusive ways:
(I) Selecting bag X and then drawing a white ball from it.
(II) Selecting bag Y and then drawing a white ball from it.

Let E1, E2 and A be three events as defined below:
E1 = Selecting bag X
E2 = Selecting bag Y
A = Drawing a white ball

We know that one bag is selected randomly.

∴ PE1=12 PE2=12 PA/E1=25PA/E2=46=23Using the law of total probability, we getRequired probability = PA=PE1PA/E1+ PE2PA/E2 =12×25+12×23 =15+13 =3+515=815

Q24.

Answer :

b) 8179 can be obtained from throw of two dice in only 4 cases as given below:3, 6, 4, 5, 5, 4, 6, 3⇒Pgetting 9 =436=19 Pnot getting 9=3236=89Now,PB is winning = Pgetting 9 in 2nd throw + Pgetting 9 in 4th throw+Pgetting 9 in 6th throw + …=89×19+89×89×89×19+ … =8811+6481+64812+ …=881×11-6481=881×8117=817

Q25.

Answer :

P(53 Sundays in a leap year) = 27

P(53 Sundays in a non-leap year) = 17

There will be 24 leap years in the 22nd century, i.e. from the year 2201 to 2200, we will have 24 leap years.

∴ P(leap year) = 24100

P(non-leap year) = 76100

Now,

P(53 Sundays) = P(leap year)×P(53 Sundays in a leap year)
+ P(non-leap year)×P(53 Sundays in a non-leap year)

=24100×27+76100×17=48700+76700=124700=31175

Q26.

Answer :

Number divisible by 6 between 1 to 100 = 16
Number divisible by 8 between 1 to 100 = 12
Number divisible by 6 and 8 between 1 to 100 = 4
Number divisible by 24 between 1 to 100 = 4

P(number divisible by 6 or 8) = P(number divisible by 6) + P(number divisible by 8) – P(number divisible by 6 and 8)

=16100+12100-4100=24100=625
P(number divisible by 6 or 8 but not by 24) = P(number divisible by 6 or 8) – P(number divisible by 24)

=625-4100=625-125=525=15

MEAN & VARIANCE OF RANDOM VARIABLES

Page 32.13 Ex. 32.1

Q1.

Answer :

(i) P (X = 3) + P (X = 2) + P (X = 1) + P (X = 0) + P (X = -1)
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05
=1.05 > 1
It is not the probability distribution of random variable X.

(ii) P (X = 0) + P (X = 1) + P (X = 2)
= 0.6 + 0.4 + 0.2
= 1.2 > 1
It is not the probability distribution of random variable X.

(iii) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)
= 0.1 + 0.5 + 0.2 + 0.1 + 0.1
= 1
It is the probability distribution of random variable X.

(iv) P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)
= 0.3 + 0.2 + 0.4 + 0.1
= 1
It is the probability distribution of random variable X.

Q2.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = -2) + P (X = -1) + P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒0.1+k+0.2+2k+0.3+k=1⇒4k+0.6=1⇒k=0.44=0.1

Q3.

Answer :

(i) Since the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

⇒a+3a+5a+7a+9a+11a+13a+15a+17a=1⇒81a=1⇒a=181

(ii) P (X < 3)

=PX=0+PX=1+PX=2=181+381+581=981=19

P (X ≥ 3)

PX=3+PX=4+PX=5+PX=6+PX=7+PX=8=781+981+1181+1381+1581+1781=7281=89

P (0 < X < 5)
PX=1+PX=2+PX=3+PX=4=381+581+781+981=2481=827

Q4.

Answer :

(i) We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) = 1

⇒3c3+4c-10c2+5c-1=1⇒3c3-10c2+9c-2=0⇒c-13c2-7c+2=0⇒c-13c-1c-2=0⇒c= 13, 1, 2Neglecting 1 and 2 as individual probability should not be greater than one

(ii) P (X < 2)

=PX=0+PX=1=3c3+4c-10c2=19+43-109=1+12-109=39=13

(iii) P (1 < X ≤ 2)

=PX=2=5c-1=53-1=5-33=23

Q5.

Answer :

Let P (X = x3) = k. Then,
P (X = x₁) = k₂

P (X = x₂) = k₃

P (X = x₄) = k₅

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = x₁) + P (X = x₂) + P (X = x₃) + P (X = x₄) = 1

⇒k2+k3+k+k5=1⇒15k+10k+30k+6k30=1⇒61k30=1⇒k=3061

Now,

Q6.

Answer :

Let P (X = 0) = k. Then,
P (X = 0) = P (X > 0) = P (X < 0)
⇒P (X > 0) = k
P (X < 0) = k

∴ P (X = 0) + P (X > 0) + P (X < 0) = 1

⇒k+k+k=1⇒k=13

Now,
P (X < 0) = k

⇒PX=-1+PX=-2+PX=-3=k⇒3PX=-1=k ∵ PX=-1=PX=-2=PX=-3⇒PX=-1=k3⇒PX=-1=13×13=19∴ PX=-1=PX=-2=PX=-3=19Similarly, PX>0=k⇒PX=1=PX=2=PX=3=19

Thus, the probability distribution is given by

Q7.

Answer :

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=48C252C2=22562652=188221PX=1=P1 ace=4C1×48C152C2=1921326=32221PX=2=P2 aces=4C252C2=61326=1221

Thus, the probability distribution of X is given by

Q8.

Answer :

Let X denote the number of heads in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=PTTT=18, PX=1=PHTT or TTH or THT=38PX=2=PHTH or THH or HHT=38, PX=3=PHHH=18

Thus, the probability distribution of X is given by

Q9.

Answer :

Let X denote the number of aces in a sample of 4 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno ace=48C452C4PX=1=P1 ace=4C1×48C352C4PX=2=P2 aces=4C2×48C252C4PX=3=P3 aces=4C3×48C152C4PX=4=P4 aces=4C452C4

Thus, the probability distribution of X is given by

Page 32.14 Ex. 32.1

Q10.

Answer :

Let X denote the number of red balls in a sample of 3 balls drawn from a bag containing 4 red and 6 black balls. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno red ball=6C310C3=20120=16PX=1=P1 red ball=4C1×6C210C3=60120=12PX=2=P2 red balls=4C2×6C110C3=36120=310PX=3=P3 red balls=4C310C3=4120=130

Thus, the probability distribution of X is given by

Q11.

Answer :

Let X denote the number of defective mangoes in a sample of 4 mangoes drawn from a bag containing 5 defective mangoes and 15 good mangoes. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective mango=15C420C4=13654845=91323PX=1=P1 defective mango=5C1×15C320C4=22754845=455969PX=2=P2 defective mangoes=5C2×15C220C4=10504845=70323PX=2=P3 defective mangoes=5C3×15C120C4=1504845=10323PX=3=P4 defective mangoes=5C420C4=54845=1969

Thus, the probability distribution of X is given by

Q12.

Answer :

Let X denote the sum of the numbers on two die. Then, X can take the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12.
Sample space : {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1), (3, 2), (3 ,3), (3, 4), (3, 5), (3, 6)
(4, 1), (4, 2), (4 ,3), (4, 4), (4, 5), (4, 6)
(5, 1), (5, 2), (5 ,3), (5, 4), (5, 5), (5, 6)
(6, 1), (6, 2), (6 ,3), (6, 4), (6, 5), (6, 6)}

Now,
PX=2=136PX=3=236PX=4=336PX=5=436PX=6=536PX=7=636PX=8=536PX=9=436PX=10=336PX=11=236PX=12=136

Thus, the probability distribution of X is given by

Q13.

Answer :

Here, X can take the values 14, 15, 16, 17, 19, 20 and 21.
Now,
PX=14=215PX=15=115PX=16=215PX=17=315PX=18=115PX=19=215PX=20=315PX=21=115

Thus, the probability distribution of X is given by

Q14.

Answer :

Let X denote the number of defective bolts in a sample of 4 bolts drawn from a bag containing 5 defective bolts and 20 good bolts. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective bolts=20C425C4=484512650=9692530PX=1=P1 defective bolt=5C1×20C325C4=570012650=114253PX=2=P2 defective bolts=5C2×20C225C4=190012650=38253PX=3=P3 defective bolts=5C3×20C125C4=20012650=4253PX=4=P4 defective bolts=5C425C4=512650=12530

Thus, the probability distribution of X is given by

Q15.

Answer :

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=4852×4852=12×1213×13=144169PX=1=P1 ace=452×4852=2×1213×13=24169PX=2=P2 aces=452×452=1×113×13=1169

Thus, the probability distribution of X is given by

Q16.

Answer :

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno kings=4852×4852=12×1213×13=144169PX=1=P1 king=452×4852=2×1213×13=24169PX=2=P2 kings=452×452=1×113×13=1169

Thus, the probability distribution of X is given by

Q17.

Answer :

Let X denote the number of aces in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno ace=4852×4751=22562652=188221PX=1=P1 ace=452×4851+4852×451=3842652=32221PX=2=P2 aces=452×351=122652=1221

Thus, the probability distribution of X is given by

Q18.

Answer :

Let X denote the number of white balls in a sample of 3 balls drawn from a bag containing 4 white and 6 red balls. Then, X can take the values 0, 1, 2 and 3.
Now,

PX=0=Pno white ball=6C310C3=20120=16PX=1=P1 white ball=4C1×6C210C3=60120=12PX=2=P2 white balls=4C2×6C110C3=36120=310PX=3=P3 white balls=4C310C3=4120=130

Thus, the probability distribution of X is given by

Q19.

Answer :

It is given that Y denotes the number of times a total of 9 appears on throwing the pair of dice.

When the dice is thrown 2 times, the possibility of getting a total of 9 is possible only for the given combinations:

(3, 6) (4, 5) (5, 4) (6, 3)

So, the total number of outcomes is 36 and the total number of favourable outcomes is 4.

Probability of getting a total of 9 = 436=19

Probability of not getting a total of 9 = 1-19=89

If Y takes the values 0, 1 and 2, then

PY=0=89×89=6481PY=1=19×89+89×19=1681PY=2=19×19=181

Thus, the probability distribution of X is given by

Q20.

Answer :

Let X denote the number of defective items in a sample of 4 items drawn from a bag containing 5 defective items and 20 good items. Then, X can take the values 0, 1, 2, 3 and 4.
Now,
PX=0=Pno defective item=20C425C4=484512650=9692530PX=1=P1 defective item=5C1×20C325C4=570012650=114253PX=2=P2 defective items=5C2×20C225C4=190012650=38253PX=3=P3 defective items=5C3×20C125C4=20012650=4253PX=4=P4 defective items=5C425C4=512650=12530

Thus, the probability distribution of X is given by

Q21.

Answer :

Let X denote the number of hearts in a sample of 3 cards drawn from a well-shuffled deck of 52 cards. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=Pno heart=3952×3952×3952=2764PX=1=P1 heart=1352×3952×3952+3952×1352×3952+3952×3952×1352=2764PX=2=P2 hearts=1352×1352×3952+3952×1352×1352+1352×3952×1352=964PX=3=P3 hearts=1352×1352×1352=164

Thus, the probability distribution of X is given by

Q22.

Answer :

Let X denote the number of blue balls in a sample of 3 balls drawn from a bag containing 4 red and 3 blue balls. Then, X can take values 0, 1, 2 and 3.
Now,
PX=0=Pno blue ball=47×47×47=64343PX=1=P1 blue ball=37×47×47+47×37×47+47×47×37=144343PX=2=P2 blue balls=37×37×47+47×37×37+37×47×37=108343PX=3=P3 blue balls=37×37×37=27343

Thus, the probability distribution of X is given by

Q23.

Answer :

Let X denote the number of spades in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,

PX=0=Pno spade=39C252C2=7411326=1934PX=1=P1 spade=13C1×39C152C2=5071326=1334PX=2=P2 spades=13C252C2=781326=117

Thus, the probability distribution of X is given by

Q24.

Answer :

Let X denote the event of getting a number less than 3 (1 or 2) on throwing the die. Then, X can take the values 0, 1 and 2.
Now,
PX=0=1636=49PX=1=1636=49 PX=2=436=19

Thus, the probability distribution of X is given by

Q25.

Answer :

The possible values of X are 0, 1 and 2, i.e. no black ball, 1 black ball and 2 black balls.
Yes, X is a random variable.
A random variable is a real valued function having domain as the sample space associated with a random experiment.

Q26.

Answer :

Given: X = Number of heads − Number of tails

Therefore, the possible values of X are :
−6, −4, −2, 0, 2, 4, 6

Q27.

Answer :

Let X denote the number of defective bulbs in a sample of 2 bulbs drawn from a lot of 10 bulbs containing 3 defectives and 7 non-defectives.Then X can take values 0, 1, 2.
Now,
PX=0=Pno defective bulb=7C210C2=715PX=1=P1 defective bulb=3C1×7C110C2=715PX=2=P2 defective bulbs=3C210C2=115

Thus, the probability distribution of X is given below,

Page 32.34 Ex. 32.2

Q2.

Answer :

Let X denote the number of heads in three tosses of a coin.Then, X can take values 0, 1, 2 and 3.
Now,
PX=0=PTTT=18, PX=1=PHTT or TTH or THT=38PX=2=PHTH or THH or HHT=38, PX=3=PHHH=18

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=32Variance=∑pixi2-Mean2 =3-322 =34

Q3.

Answer :

Let X denote the number of kings in a sample of 2 cards drawn from a well-shuffled pack of 52 playing cards. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno king=48C252C2=11281326=188221PX=1=P1 king=4C1×48C152C2=1921326=32221PX=2=P2 kings=4C252C2=61326=1221

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=34221Variance=∑pixi2-Mean2            =36221-342212          =7956-115648841          =680048841          =4002873

Q4.

Answer :

Let X denote the number of tails in three tosses of a coin. Then, X can take the values 0, 1, 2 and 3.
Now,
PX=0=PHHH=18, PX=1=PTHH or HHT or HTH=38PX=2=PTTH or THT or HTT=38, PX=3=PTTT=18

Thus, the probability distribution of X is given by

Computation of mean and step deviation

Mean=∑pixi=32Variance=∑pixi2-Mean2  =3-322 =34Step Deviation=Variance =34    =0.87

Q5.

Answer :

Let X denote the number of bad eggs in a sample of 3 eggs drawn from a lot containing 2 bad eggs and 10 good eggs. Then, X can take the values 0, 1 and 2.
Now,
PX=0=Pno bad egg=10C312C3=120220=611PX=1=P1 bad egg=2C1×10C212C3=90220=922PX=2=P2 bad eggs=2C2×10C112C3=10220=122

Thus, the probability distribution of X is given by

Computation of mean

Mean=∑pixi=12

Q6.

Answer :

Let X denote the event of getting twice the number. Then, X can take the values 1, 2, 3, 4, 5 and 6.

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2.5Variance=∑pixi2-Mean2=8.4-6.25=2.15

Q7.

Answer :

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTT, HTTT, THHH, …

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when a coin is tossed 4 times, we can get minimum 0 and maximum 4 strings.)

Now,
PX=0=P0 head=116PX=1=P1 head=416PX=2=P2 heads=616PX=3=P3 heads=416PX=4=P4 heads=116

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2Variance=∑pixi2-Mean2  =5-4  =1

Q8.

Answer :

Let X denote the event of getting twice the number. Then, X can take the values 2, 4, 6, 8, 10 and 12.

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=7Variance=∑pixi2-Mean2  =60.7-49 =11.7

Q9.

Answer :

Let X be 1 for the appearance of odd numbers 1, 3 or 5 on the die. Then,
PX=1=36=12

Let X be 3 for the appearance of even numbers 2, 4 or 6 on the die. Then,
PX=3=36=12

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2Variance=∑pixi2-Mean2  =5-4  =1

Q10.

Answer :

If a coin is tossed 4 times, then the possible outcomes are:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH and TTTT .

For the longest string of heads, X can take the values 0, 1, 2, 3 and 4.
(As when the coin is tossed 4 times, we can get maximum 4 and minimum 0 strings.)

Now,
PX=0=P0 head=116PX=1=P1 head=716PX=2=P2 heads=516PX=3=P3 heads=216PX=4=P4 heads=116

Thus, the probability distribution of X is given by

Computation of mean and variance

Mean=∑pixi=2716=1.7Variance=∑pixi2-Mean2   =6116-729256  =247256 =0.9

Page 32.36 (Very Short Answers)

Q1.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = -2) + P (X = -1) + P (X = 0) + P (X = 1) = 1

⇒1-a4+1+2a4+1-2a4+1+a4=1⇒44=1⇒1=1Now, 0≤1-a4≤1⇒ 0≤1-a≤4⇒ -1≤-a≤3⇒ -3≤a≤1 …10≤1+a4≤1⇒ 0≤1+a≤4⇒ -1≤a≤3 …20≤1-2a4≤1⇒ 0≤1-2a≤4⇒ -1≤-2a≤3⇒ -32≤a≤12 …30≤1+2a4≤1⇒ 0≤1+2a≤4⇒ -1≤2a≤3⇒ -12≤a≤32 …4From 1, 2, 3 and 4, we get-12≤a≤12

Q2.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒2k4+3k2-5k3+2k-3k2+3k-1=1⇒2k4-5k3+5k=2⇒2k4-5k3+5k-2=0⇒k-1k-22k2+k-1=0⇒k-1k-22k-1k+1=0⇒k=-1 , 12, 1, 2Neglecting -1 , 1 and 2 as they give the value of probability negative or greater than 1

∴ k =12

Q3.

Answer :

A cubical die can show 1, 2, 3, 4, 5 or 6 on its face.

Mean = ∑pixi = 16+26+36+46+56+66=216=3.5

Q4.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒2k+4k+3k+k=1⇒10k=1⇒k=110=0.1

Q5.

Answer :

Mean = ∑pixi = 14+28+158=2+2+158=198

Q6.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒c+2c+4c+4c=1⇒11c=1⇒c=111Now,PX≤2=PX=1+PX=2=110+210=311

Page 32.37 (Very Short Answers)

Q7.

Answer :

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 1

⇒k+2k+3k+4k=1⇒10k=1⇒k=110Now, PX≥3=PX=3+PX=4=310+410=710

Page 32.37 (Multiple Choice Questions)

Q1.

Answer :

(d) 181
We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

⇒a+3a+5a+7a+9a+11a+13a+15a+17a=1⇒81a=1⇒a=181

Q2.

Answer :

(b) 0.77

E = {X is a prime number} = {2, 3, 5, 7}

PE =P2+P3+P5+P7= 0.62F= {X<4} = {X=1,2,3} PF= P1+P2+P3=0.5Now,E∩F = {2,3}PE∩F=P2+P3=0.35PE∪F=PE+PF-PE∩F=0.62+0.50-0.35=0.77

Q3.

Answer :

(d) 0.4

Let:
P(X = 0) = m
P(X = 1) = k.

Now,
P(X = 3) = 2k

Mean = ∑pixi

0+k+0.6+6k=1.3⇒7k=1.3-0.6⇒k=0.77=0.1

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒m+0.1+0.3+0.2=1⇒m+0.6=1⇒m=0.4

Q4.

Answer :

(a) 1/10

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8) = 1

⇒0+2p+2p+3p+p2+2p2+7p2+2p=1⇒10p2+9p-1=0⇒10p-1p+1=0⇒p=110 or -1 Neglecting -1 as the value of the probabilitiy cannot be negative

Q5.

Answer :

(d) 1/8, 3/4

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

⇒k+3k+3k+k=1⇒8k=1⇒k=18

Now,

Mean=∑pixi=32Variance=∑pixi2-Mean2=248-322=248-94=24-188=68=34

Page 33.12 Ex. 33.1

Q1.

Answer :

Let X denote the number of defective items in a sample of 8 items. Then, X follows a binomial distribution with n = 8, p = (probability of getting a defective item) = 0.06 and q = 1-p=0.94

P(X=r) =Cr8 (0.06)r (0.94)8-r, r=0,1,2,3,…8The required probability = Probability of not more than one defective item = P (X≤1) = P(X=0)+P(X=1) = C08 (0.06)0 (0.94)8-0+ C18 (0.06)1 (0.94)8-1 = (0.94)8 +8(0.06)(0.94)7 = (0.94)7 0.94+0.48 = 1.42(0.94)7

Q2.

Answer :

Let X denote the number of heads in 5 tosses.X follows a binomial distribution with n =5; p = probability of getting a head = 12and q = 1-p =12P(X=r) = Cr512r125-r, r =0,1,2…5The required probability = P(getting at least 3 heads)= P(X≥3) = P(X=3)+P(X=4)+P(X=5) = C35123125-3+ C45121125-1 + C55125125-0
=10125 + 5125 + 1125 = 125(10+5+1) = 125×16 = 12

Q3.

Answer :

Let X denote the number of tails when a coin is tossed 5 times.X follows a binomial distribution with n =5; p = 12; q =1-p =12Then P(X=r) =Cr512r 12n-r = Cr5125 The required probability = P(X=odd) =P(X=1)+P(X=3)+P(X=5)=C15125 +C35125 +C55125 = 125 [5+10+1] = 1632=12

Q4.

Answer :

Let X be the number of successes in 6 throws of the two dice.

Probability of success = Probability of getting a total of 9
= Probability of getting (3,6), (4,5), (5,4), (6,3) out of 36 outcomes

p=436=19, q =1-p =89and n = 6X follows a binomial distribution with n=6, p=19 and q=89P(X=r) = Cr619r 896-r The required probability = Probability of at least 5 successes= P(X≥5) = P(X=5)+P(X=6)=C56 195 896-5 +C66 196 896-6 = 6(8)+196=4996

Q5.

Answer :

Let X denote the number of heads when 6 coins are tossed.Then, X follows a binomial distribution with n =6; p = 12and q =12;P(X=r) = Cr612r126-r = Cr6125, r =0,1,2…..6

The required probability= P( X≥3) = P(X=3)+P(X=4)+P(X=5)+P(X=6)= C36126+ C46126+ C56126+ C66126=(20 + 15++6+1)164=2132

Q6.

Answer :

Let X be the probability of getting 4 in two tosses of a fair die.
X follows a binomial distribution with n =2; p =16 and q =56;
P(X=r) = Cr216r562-rProbability of getting 4 at least once = P(X≥1) = 1-P(X=0) = 1- C02160562-0 = 1-2536= 1136

Q7.

Answer :

Let X be the number of heads that appear when a coin is tossed 5 times.

X follows a binomial distribution with n =5 and p= q=12
P(X=r) = Cr512r125-r = Cr5125

P (head appears an even number of times)=P(X=0)+P(X=2)+P(X=4)= C05125+C25125+C45125 = 1+10+525= 1632=12

Q8.

Answer :

Let X be number of times the target is hit. Then, X follows a binomial distribution with n =7, p =14and q =34

P(X=r) = Cr714r347-rP( hitting the target at least twice)=P(X≥2) = 1-P(X=0) +P(X=1)= 1-C07140347-0-C17141347-1 = 1-347 – 714346= 1-116384(2187+5103) = 1-36458192=45478192

Q9.

Answer :

Let X be the number of busy calls for 6 randomly selected telephone numbers.

X follows a binomial distribution with n =6 ;p = one out of 15 = 115and q = 1415
P(X=r) = Cr6115r14156-rProbability that at least 3 of them are busy = P(X≥3) = 1-{P(X=0)+P(X=1)+P(X=2)}= 1 -C06115014156-0+C16115114156-1+C26115214156-2

=1-14156+61514155+11514154

Q10.

Answer :

Let X denote the number of successes, i.e. of getting 5 or 6 in a throw of die in 6 throws.

Then, X follows a binomial distribution with n =6;
p = of getting 5 or 6 = 16+16=13; q = 1-p = 23;P(X=r) = Cr613r236-rP(X ≥4) = P(X=4)+P(X=5)+P(X=6)=C46134236-4+C56135236-5+C66136236-6 = 136(60+12+1) =73729

Q11.

Answer :

Let X be the number of heads in tossing 8 coins.
X follows a binomial distribution with n =8; p = 12and q = 12;
P(X=r) = Cr812r128-r = Cr8128Probability of obtaining at least 6 heads = P(X≥6) = P(X=6)+P(X=7)+P(X=8)= C68128 + C78128 + C88128 = 12828+8+1 = 37256

Q12.

Answer :

Let X denote the number of spade cards when 5 cards are drawn with replacement. Because it is with replacement,

X follows a binomial distribution with n = 5; p =1352=14; q = 1-p = 34

P(X=r) =Cr514r345-r

(i) P(All cards are spades) = P(X=5) =C55145340 = 11024(ii) P(only 3 cards are spades) = P(X=3) =C35143342 = 1102490 = 45512iii P(none is a spade) = P(X=0)=C05140345 = 2431024

Q13.

Answer :

Let X be the number of white balls drawn when 4 balls are drawn with replacement.
X follows binomial distribution with n = 4.

p = Probability for a white ball = No of white ballsTotal no. of balls = 520=14and q = 1-p = 34P(X=r) = Cr414r344-r(i) Prob that none is white = P(X=0) =C04140344-0= 81256ii Prob that all are white = P(X=4) = C44144344-4= 1256iii Prob that any two are white = P(X=2) = C24142344-2= 54256=27128

Page 33.13 Ex. 33.1

Q14.

Answer :

Let X be the variable representing number on the ticket bearing a number divisible by 10 out of the 5 tickets drawn.
Then, X follows a binomial distribution with n =5;

p = Probability of getting a ticket bearing number divisible by 10. p = 1100(10) = 110; q=910;P(X=r) = Cr5110r9105-rProbability that all the tickets bear numbers divisible by 10= P(X=5) =C5511059105-5=11059100=1105

Hence, required probability is 1105

Q15.

Answer :

Let X be the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

As this is with replacement, X follows a binomial distribution with n = 4;
p = probabilty that a ball randomly drawn bears digit 0 =110;q=1-p = 910;

P(X=r) = Cr4110r9104-r P(none bears the digit 0) = P(X=0) = C0411009104-0 = 9104

Q16.

Answer :

Let X denote the number of defective items in a sample of 10 items.

X follows a binomial distribution with n =10; p = probability of defective items = 5% = 0.05 ;q = 1-p =0.95

P(X=r) =Cr10(0.05)r(0.95)10-rProbability ( sample of 10 items will include not more than one defective item)= P(X≤1) = P(X=0)+P(X=1) =C010(0.05)0(0.95)10-0+C110(0.05)1(0.95)10-1 =( 0.95)9(0.95+0.5)= 1.45(0.95)9

=192092920; (∵ 1.45 = 2920and 0.95 = 1920)

Q17.

Answer :

Let X be the number of bulbs that fuse after 150 days.

X follows a binomial distribution with n = 5, p =0.05 and q =0.95
Or p = 120and q = 1920P(X=r) = Cr5120r19205-r(i) Probability (none will fuse after 150 days of use ) = P(X=0) =C05120019205-0 =19205(ii) Probability (not more than 1 will fuse after 150 days of use) = P(X≤1) = P(X=0)+P(X=1) =19205+5C1120119205-1 = 192041920+520 = 6519204
(iii) Probability more than one will fuse after 150 days of use = P(X>1) = 1-P(X≤1)= 1-6519204 ∵P(X≤1)=6519204 iv Probability at least one will fuse after 150 days of use = P(X≥1)= 1-P(X=0) = 1-19205 ∵P(X=0=19205

Q18.

Answer :

Let X be the number of people that are right-handed in the sample of 10 people.

X follows a binomial distribution with n = 10,
p =90% =0.9 and q = 1-p = 0.1P(X=r) = Cr10(0.9)r(0.1)10-rProbability that at most 6 are right-handed = P(X≤6)= P(X=0)+ P(X=1)+ P(X=2)+ P(X=3)+ P(X=4)+ P(X=5)+ P(X=6)= 1-{P(X=7)+ P(X=8)+ P(X=9)+ P(X=10)}= 1- ∑r=710Cr10(0.9)r(0.1)10-r

Q19.

Answer :

Let X denote the number of red balls drawn from 16 balls with replacement.
X follows a binomial distribution with n = 4,p = 516, q =1-p = 1116

P(X=r) = Cr4516r11164-rP(One ball is red)=P(X=1) = C14516111164-1= 451611163=5411163

Q20.

Answer :

Let X denote the number of white balls when 2 balls are drawn from the bag.

X follows a distribution with values 0,1 or 2.
P(X=0) = P(All balls non-white) = C27C29= 4272=2136P(X=1) = P ( Ist ball white and IInd ball non-white )  =C17C12C29= 1436P(X=2) = P(Both balls white) = C22C29= 136It can be shown in tabular form as follows.X 0 1 2P(X) 2136 1436 136

Q21.

Answer :

As three balls are drawn with replacement, the number of white balls, say X, follows binomial distribution with n =3
p =37 and q =47P(X=r) = Cr337r473-r, r=0,1,2,3P(X=0) =C03370473-0 P(X=1) = C13371473-1 P(X=2) = C23372473-2 P(X=3) =C33373473-3X 0 1 2 3P(X) 64343 144343 108343 27343

Q22.

Answer :

Let X be the number of doublets in 4 throws of a pair of dice.

X follows a binomial distribution with n =4,
p = No of getting (1,1)(2,2)…(6,6) = 636=16q =1-p = 56P(X=r) = Cr416r564-r, r= 0,1,2,3,4P(X=0) = C04160564-0P(X=1) = C14161564-1P(X=2) = C24162564-2P(X=3) = C34163564-3P(X=4) = C44164564-4The distribution is as follows.X 0 1 2 3 4 P(X) 6251296 5001296 1501296 201296 11296

Q23.

Answer :

Let X be the number of 6 in 3 tosses of a die.

Then X follows a binomial distribution with n =3.

p = 16, q=1-p =56P(X=r) = Cr316r563-r, r=0,1,2,3P(X=0) = C03160563-0P(X=1) = C13161563-1P(X=2) = C23162563-2P(X=3) = C33163563-3

Hence, the distribution of X is as follows.X 0 1 2 3P(X) 125216 75216 15216 1216

Q24.

Answer :

Let X = number of heads in 5 tosses. Then the binomial distribution for X has n =5, p =12 and q =12
P(X=r) = Cr512r125-r, r = 0,1,2,3,4,5 = Cr525Substituting r =0,1,2,3,4,5 we get the following probability disrtribution.X 0 1 2 3 4 5P(X) 132 532 1032 1032 532 132

Q25.

Answer :

Let X denote getting a number greater than 4 .

Then, X follows a binomial distribution with n=2

p = P(X>4)=P(X=5 or 6)= 16+16=13q =1-p = 23P(X=r) = Cr2 13r232-r , r=0,1,2 Substituting for r we get probability distribution of X as follows.

X 0 1 2P(X) 49 49 19

Q26.

Answer :

Let X be the number of rupees the man wins.

First, let us assume that he gets head in the first toss.Probability would be 12. Also, he wins Re.1.The second possibility is that he gets a tail in the first toss. Then he tosses again.Suppose he gets a head in the second toss.Then, he wins Re 1 in the second toss but loses Re 1 in the first toss. So, the money won =Rs. 0Probability for winning Rs .0 =Getting tail in Ist toss x Getting head in IInd toss=12×12=14The third possibility is getting tail in the Ist toss and also tail in the IInd toss.Then, the money that he would win = -2 (As he loses Rs. 2)Probability for the third possibility = 12×12=14Probability distribution of X is as follows.X 1 0 -2P(X) 12 1 4 14

Q27.

Answer :

Let X denote the occurrence of 3,4 or 5 in a single die. Then, X follows binomial distribution with
n=5.
Let p=probability of getting 3,4, or 5 in a single die .

p = 36=12

q=1-12=12 P(X=r) = Cr512r125-rP(at least 3 successes) =P(X≥3) = P(X=3)+P(X=4)+P(X=5)= C35123125-3+C45124125-4+C55125125-5= C35+C45+C5525= 12

Q28.

Answer :

Let X denote the number of defective items in the items produced by the company.
Then, X follows binomial distribution with n = 8.

p = 10%=110q= 1-p = 910Hence, the distribution is given byP(X=r) = Cr8110r9108-rProb of getting 2 defective items =P(X=2) =C2811029108-2 = 28 x 96108

Q29.

Answer :

(i) Let p denote the probability of drawing a heart from a deck of 52 cards. So,

p=1352=14and q=1-q=1-14 = 34

Let the card be drawn n times. So, binomial distribution is given by: P(X=r)=Crnprqn-r
Let X denote the number of hearts drawn from a pack of 52 cards.
We have to find the smallest value of n for which P(X=0) is less than 14
P(X=0) < 14
C0n14034n-0<1434n<14Put n=1, 341 not less than 14 n=2, 342 not less than 14 n=3, 343 not less than 14So, smallest value of n=3
Therefore card must be drawn three times.

(ii) Given the probability of drawing a heart > 34
1 – P(X=0) > 34
1- C0n14034n-0>341-34n>341-34>34n14>34n

For n=1, 341 not less than 14 n=2, 342 not less than 14 n=3, 343 not less than 14 n=4, 344 not less than 14 n=5, 345 not less than 14
So, card must be drawn 5 times.

Q30.

Answer :

Let k be the number of desks and X be the number of graduate assistants in the office.
therefore, X=8, p=12, q=12
According to the given condition,
PX≤k>90%⇒PX≤k>0.90⇒PX>k<0.10⇒P(X=k+1, k+2,….8)<0.10
Therefore, P(X > 6) = P(X=7 or X=8)
C78128 + C88128 =0.04

Now, P(X > 5) = P(X = 6, X = 7 or X = 8) = 0.15
P(X > 6) < 0.10

So, if there are 6 desks then there is at least 90% chance for every graduate to get a desk.

Q31.

Answer :

Let X be the number of heads in tossing the coin 8 times.
X follows a binomial distribution with n =8
p =12 and q=12Hence, the distribution is given by ∴P(X=r) = Cr812r128-r, r=0,1,2,3,4,5,6,7,8Required probability =P(X≥6) = P(X=6)+P(X=7)+P(X=8)= C68+ C78+ C8828= 28+8+1256=37256

Q32.

Answer :

Let X denote the number of heads obtained in tossing 6 coins.
Then, X follows a binomial distribution with n=6,

p = 12 and q=12Hence, the distribution is given byP(X=r) =Cr612r126-r , r=0,1,2,3,4,5,6= Cr626i P(getting 3 heads) =P(X=3) =C3626= 2064=516ii P(getting no head) =P(X=0) =C0626= 126 = 164iii P(getting at least 1 head) = P(X≥1) =1- P(X=0) = 1-164=6364

Page 33.14 Ex. 33.1

Q33.

Answer :

Let X denote the number of tubes that function for more than 500 hours.
Then, X follows a binomial distribution with n =4.
Let p be the probability that the tubes function more than 500 hours.

Here, p=0.2, q =0.8Hence, the distribution is given byP(X=r) =Cr4(0.2)r(0.8)4-r, r= 0,1,2,3,4Therefore, required probability = P(X=3) = 4(0.2)3(0.8) = 0.0256

Q34.

Answer :

Let X denote the number of components that survive shock.
Then, X follows a binomial distribution with n = 5.
Let p be the probability that a certain kind of component will survive a given shock test.
∴p = 34and q=14Hence, the disrtibution is given byP(X=r) = Cr534r145-r, r=0,1,2,3,4,5i P(exactly 2 will survive)=P(X=2) =C25342145-2 = 10×91024= 0.0879

ii P(atmost 3 will survive) = P(X≤3)= P(X=0)+ P(X=1)+ P(X=2) +P(X=3)=C05340145-0+C15341145-1+C25342145-2+C35343145-3= 145+534144+10342143+10343142= 1+15+90+2701024= 3761024= 0.3672

Q35.

Answer :

Let X be the number of bombs that hit the target.
Then, X follows a binomial distribution with n = 6
Let p be the probability that a bomb dropped from an aeroplane will strike the target.

∴ p =0.2 and q =0.8Hence, the distribution is given byP(X=r) =Cr60.2r0.86-r(i) P(exactly 2 will strike the target)=P(X=2) = C26(0.2)2(0.8)4=15(0.04)0.4096=0.2458(ii) P(at least 2 will strike the target) = P(X≥2) = 1-[P(X=0)+P(X=1)]= 1- (0.8)6-6(0.2)(0.8)5 = 1-0.2621-0.3932= 0.3447

Q36.

Answer :

Let X be the number of mice that contract the disease .
Then, X follows a binomial distribution with n =5.
Let p be the probability of mice that contract the disease.
∴ p=0.4 and q = 0.6Hence, the distribution is given byP(X=r)=Cr50.4r0.65-r , r=0,1,2,3,4,5i P(X=0) =C050.400.65-0= (0.6)5 = 0.0778ii P(X>3) =P(X=4)+P(X=5)=C450.440.65-4+C550.450.65-5=0.0768+0.01024= 0.08704

Q37.

Answer :

Let X denote the number of successes in 6 trials.

It is given that successes are twice the failures. ⇒p=2q p+q=1⇒3q=1⇒q=13∴ p=1-13=23 n =6Hence, the distribution is given by

P(X=r)=Cr623r136-r , r=0,1,2…..6P(atleast 4 successes)=P(X≥4) = P(X=4)+P(X=5)+P(X=6)C46234136-4+C56235136-5+C66236136-6= 15(24)+6(32)+6436= 240+192+64729= 496729

Q38.

Answer :

Let X denote the number of machines out of service during a day.Then, X follows a binomial distribution with n=20 Let p be the probability of any machine out of service during a day.∴p= 0.02 and q =0.98 Hence, the distribution is given byP(X=r) =Cr20 0.02r0.9820-r , r=0,1,2…..20∴P(exactlly 3 machines will be out of the service on the same day)=P(X=3) =C3 200.0230.9820-3 = 1140(0.000008)(0.7093)= 0.006469

Q39.

Answer :

Let X be the number of students that graduate from among 3 students.
Let p=probability that a student entering a university will graduate.

Here , n =3, p=0.4 and q = 0.6
Hence, the distribution is given by
P(X=r)=Cr30.4r0.63-r , r=0,1,2,3

i P(X=0) = q3 =0.216ii P(X=1) = 3(0.4)(0.36) =0.432iii P(X =3) = p3 = 0.064

Q40.

Answer :

Let X be the number of defective eggs drawn from 10 eggs.
Then, X follows a binomial distribution with n =10
Let p be the probability that a drawn egg is defective.

∴p =10%=110 , q =910Hence, the distribution is given by P(X=r) =Cr10110r91010-r , r=0,1,2….10P(there is at least one defective egg) =P(X≥1) = 1-P(X=0) =1-C010110091010-0= 1-91010

=1-9101010

Q41.

Answer :

Let X denote the number of correct answers.
Then, X follows a binomial distribution with n =20

Let p be the probability of a correct answer. ⇒p= getting a head and a right answer to be true or getting a tail and a right answer to be false⇒p= 12∴ q=1- 12=12Hence, the distribution is given byP(X=r) = Cr2012r1220-r , r=0,1,2,3……20= Cr20220Probability that the student answers at least 12 questions correctly=P(X≥12) = P(X=12)+P(X=13)+…+P(X=20)= C1220+C1320+…+C2020220

Q42.

Answer :

We have n=6 and p=12∴ q = 1-p =12Hence, the distribution is given byP(X=r) =Cr612r126-r, r=0,1,2,3,4,5,6= Cr6126

P(X=r) = Cr626 By substituting r =0,1,2,3,4,5 and 6, we get the following distribution for X.X 0 1 2 3 4 5 6P(X) 164 664 1564 2064 1564 664 164

Comparing the probabilities, we get that X = 3 is the most likely outcome.

Q43.

Answer :

Let X be the number of right answers in the 5 questions.
X can take values 0,1,2…5.
X follows a binomial distribution with n =5
.p =probability of guessing right answer = 13 q =probability of guessing wrong answer = 23Hence, the distribution is given byP(X=r)=Cr5 13 r 235-r , r=0,1,2,…5∴ P(The student guesses 4 or more correct answers) = P(X≥4) = P(X=4)+P(X=5)=C45134231+C55135230 = 10+135=11243

Q44.

Answer :

Let X denote the number of times the person wins the lottery.
Then, X follows a binomial distribution with n = 50.

Let p be the probability of winning a prize.∴ p = 1100, q=1-1100=99100Hence, the distribution is given byP(X=r) = Cr501100r9910050-r , r=0,1,2 …50i P(winning at least once) = P(X≥0) = 1-P(X-0)= 1-9910050
ii P(winning exactly once) = P(X=1) =C150110019910050-1 = 129910049

iii P(winning at lease twice) = P(X≥2) = 1-P(X=0)-P(X=1)= 1-9910050-C150 ×1100×9910049 =1-9949 ×14910050

Q45.

Answer :

Let the shooter fire n times and let X denote the number of times the shooter hits the target.
Then, X follows binomial distribution with p =34and q =14 such that

PX=r=Crn 34r14n-r ⇒PX=r=Crn 3r4nIt is given that PX≥1 >0.99⇒1-PX=0 >0.99⇒1-14n>0.99⇒14n<0.01⇒4n>10.01⇒4n>100The least value of n satisfying this inequality is 4. Hence, the shooter must fire at least 4 times.

Q46.

Answer :

Suppose the man tosses a fair coin n times and X denotes the number of heads in n tosses.
As p=12 and q=12, P(X=r) =Crn12r12n-r, r=0,1,2,3….nIt is given that PX≥1 >0.9⇒1-PX=0 >0.9⇒1-C0n 12n>0.9⇒12n<110⇒2n>10⇒n=4,5,6….Hence, the man must toss the coin at least 4 times.

Q47.

Answer :

Let X be the number of heads and n be the minimum number of times that a man must toss a fair coin so that probability of X≥1 is more than 80% and X follows a binomial distribution with

p = 12, q=12P(X=r) = Crn12nWe have P(X≥1) = 1-P(X=0) =1-C0n12n =1-12nand P(X≥1) >80%1-12n>80% = 0.8012n<1-0.80 =0.20 2n>10.2 =5; We know, 22 <5 while 23>5So, n =3 So, n should be atleast 3.

Q48.

Answer :

Let p denote the probability of getting a doublet in a single throw of a pair of dice. Then,
p = 636 = 16q = 1-p=1-16=56

p= Let X be the number of getting doublets in 4 throws of a pair of dice. Then, X follows a binomial distribution with n =4,
p=16and q=56P(X=r)=Probability of getting r doubletsP(X=r) = Cr4(16)r(56)4-r; r= 0,1,2,3,4If X=0, then P(X=0) = C04(16)0(56)4-r⇒P = 564If X=1, then P = C14161564-1=23563If X=2, then P = C24162564-2=16562 If X=3, then P = C34163564-3=103163

If X=4, then P =C44(16)4(56)4-4 =164= 11296

X 0 1 2 3 4P(X) (56)4 23(56)3 25216 5324 11296

Q49.

Answer :

Let X denote the number of defective bulbs in a sample of 4 bulbs drawn successively with replacement .
Then, X follows a binomial distribution with the following parameters: n=4, p =630=15and q =45
Then, the distribution is given byP(X=r) =Cr415r454-r , r=0,1,2,3,4

P X=0=454 = 256625 P X=1 = 4151453 = 256625 PX=2= 6152452 = 96625 PX=3 = 4153451 = 16625 PX=4 =154 = 1625

X 0 1 2 3 4

P(X) 256625 256625 96625 16625 1625

Page 33.24 Ex. 33.2

Q1.

Answer :

No.

The mean of a binomial distribution is np and variance is npq.

If mean is less than its variance, then np <npq.

As both n and p are positive, we can divide both sides by np.

We get 1<q, which is not true as q <1 under all circumstances. (As p+q=1, q cannot be greater than 1)

So, the mean of a binomial distribution cannot be less than its variance.

Q2.

Answer :

It is given that mean, i.e. np = 9 and variance, i.e. npq =94

∴npqnp =94×19 = 14 Hence, q=14 and p =1-q = 34When p = 34,np = 3n4 9=3n4⇒n =12P(X=r) =Cr1234r1412-r, r=0,1,2,3,4,5,…..12

Q3.

Answer :

Given: Mean = 9 and variance = 6

∴ np =9 …(1) npq =6 …(2) Dividing eq (2) by eq (1), we get q =23and p =1-q=13As np =9, substituting the value of p, we get n3=9 or n =27P(X=r) = Cr2713r2327-r, r=0,1,2….27

Q4.

Answer :

Number of trials in the binomial distribution = 5

If p is the probability for success, then

np + npq = 4.8
Or 5p+5p (1-p) = 4.8

⇒5p + 5p-5p2 =4.8 Or p2-2p+0.96 =0By factorising, we get (p-0.8)(p-1.2) =0As p cannot exceed 1, p =0.8 or 45and q= 1-p = 15∴ P(X=r) = Cr545r155-r, r=0,1,2,….5

Q5.

Answer :

Mean, i.e. np =20 ….(1)

Variance, i.e. npq =16 ….(2)

Dividing eq (2) by eq (1), we get npqnp= 1620⇒q = 45⇒ p = 1-q ∴p= 15 As np = 20 ⇒ n = 100∴P(X=r) =Cr10015r45100-r , r=0,1,2….100

Q6.

Answer :

Given:Sum of the mean and variance= 253⇒np+npq=253 ⇒np 1+q=253 …(1)Product of the mean and variance = 503 ⇒np(npq) = 503 …(2)Dividing eq (2) by eq (1), we get np(npq)np(1+q)= 503×325⇒npq1+q=2⇒npq = 2(1+q) ⇒np(1-p) = 2(2-p)⇒np = 2(2-p)(1-p) Substituting this value in np+npq =253, we get2(2-p)(1-p) (2-p) = 253⇒ 6(4-4p+p2) = 25-25p⇒6p2+p-1 =0 ⇒(3p-1)(2p+1) =0 ⇒p =13or -12. As p cannot be negative, the only answer for p is 13.
q=1-p =23 ⇒np+npq = 253⇒n131+23 =253⇒n = 15∴P(X=r) = Cr1513r2315-r, r=0,1,2……15

Q7.

Answer :

Given that mean, i.e. np = 20 …(1)
and standard deviation, i.e. npq = 4

npq =4 ⇒npq =16 …(2)Dividing eq (2) by eq (1), we get q= 1620=45and p = 15; ∴ n = Meanp= 100 P(X=r) = Cr10015r45100-r, r=0,1,2…..100Therefore, the parameters are n=100 and p = 15

Q8.

Answer :

Total number of bolts (n) = 400 and p = prob of defective bolts = 0.1

(i) Mean = np = 400(0.1) =40

(ii) Variance = npq = 40(1-0.1) = 36

So, the standard deviation = 36 =6

Q9.

Answer :

Mean of binomial distribution, i.e. np =5

Variance, i.e. npq =103
q= VarianceMean= 23and p = 1-q= 13np = 5 ⇒ n= 15∴P(X=r) = Cr1513r2315-r, r=0,1,2……15

Q10.

Answer :

Total number of ships (n) = 500
Let X denote the number of ships returning safely to the ports.
p=910 and q =1-p =110

Mean = np = 450 and Variance =npq = 45 Mean =450 Standard deviation = 45 = 6.71

Q11.

Answer :

Given: mean =16 and variance = 8
Let n and p be the parameters of the distribution.

That is, np = 16 and npq = 8

q= npqnp=12and p = 1-q =12 np = 16 ⇒n = 32∴P(X=r) = Cr3212r1232-r, r=0,1,2….32,⇒P(X=0) =1232 P(X=1) = 321232=122 P(X≥2) = 1-P(X=0)-P(X=1)= 1-1232-1227= 1-1+32232 = 1-33232

Q12.

Answer :

Let X denote the number of successes in 8 throws.

n =8
p = probability of getting 5 or 6 =26=13 and q = 23

Mean (np) =83Variance (npq) = 169Standard deviation= Variance = 43So, mean = 2.66 and standard deviation = 1.33

Q13.

Answer :

Here, n =8
Let p be the probability of number of boys in the family.

p =12, q=12Expected number of boys =mean ⇒ np = 4

Q14.

Answer :

Here, n =10,000
Let p (the probability of getting a defective item) = 0.02
q =1-0.02 = 0.98

Mean= Expected number of defective items ⇒np =200Variance (npq) = 200(0.98) = 196Standard deviation = Variance = 14So, mean =200 and standard deviation =14

Q15.

Answer :

Here, n =3
p = probability of getting 1 or 6 =13and q =1-13=23Mean = np =1Variance = npq = 23

Q16.

Answer :

Mean(np) =3 and variance (npq)=32∴ q=12and p =1-12n=Meanp ⇒n=6

Hence, the distribution is given by P(X=r) = Cr612r126-r , r=0,1,2…6= Cr626 ∴P(X≤5) = 1-P(X=6) = 1-164=6364

Q17.

Answer :

Here, mean (np) = 4
variance (npq ) =2

∴q= 12and p =12 n=Mean p=4×2=8
Hence, the distribution is given by P(X=r) =Cr812r128-r, r=0,1,2……8P(X≥5) = P(X=5)+P(X=6)+P(X=7)+P(X=8)=128C58+C68+C78+C88 = (56+28+8+1)28= 93256

Page 33.25 Ex. 33.2

Q18.

Answer :

Mean np =43and Variance npq = 89∴q = 23and p =1-23= 13Therefore, n = Meanp = 4
Hence, the distribution is given by P(X=r) = Cr413r234-r, r=0,1,2,3,4P(X≥1) = 1-[P(X=0)] = 1-234 = 81-1681= 6581

Q19.

Answer :

Given that n = 6
The sum of mean and variance of a binomial distribution for 6 trials is 103.
⇒6p+6pq = 103⇒18p+18p(1-p) =10 ⇒18p2-36p+10 =0⇒(3p-1)(6p-10) =0 ⇒ p =13or 53p=53 (Neglected as it is greater than 1)∴ p=13⇒q= 1-p = 23Hence, the distribution is given by P(X=r) = Cr613r236-r, r=0,1,2…..6

Q20.

Answer :

Let X be the number of times a doublet is obtained in four throws.

Then, p = probability of success in one throw of a pair of dice =636=16
and q = 56; n = 4P(X=r) = Cr416r564-r, r= 0,1,2,3,4As n =4 and p = 16, mean = np = 46= 23
∴ P(X=r) = Cr456r16n-r, r=0,1,2,3,4The distribution is as follows:X 0 1 2 3 4P(X) 564 2064 15064 50064 164

Q21.

Answer :

Total number of outcomes when two dice are thrown = 6 × 6, i.e. 36

Let X be the number of doublets in three throws of a pair of dice.
Then, X follows a binomial distribution with n = 3
p = P(Getting a doublet in three throws) =636=16and q = 56∴P(X=r) =Cr316r563-r, r=0,1,2,3Mean np = 316 = 12
The distribution is as follows:X 0 1 2 3P(X) 563 3161563-1 3162563-2 163 125216 75216 15216 1216

Page 33.25 (Very Short Answers)

Q1.

Answer :

n= 20 , q =0.75

⇒p = 1-q = 0.25Mean = np = 20(0.25) =5Thus, mean =5

Q2.

Answer :
Mean =5 and Variance= 4⇒np =5 and npq =4 ⇒ q= 0.8 ⇒p =1-q =0.2& np = n(0.2) =5 (given)⇒n = 50.2= 25

Q3.

Answer :
It is given that the binomial distribution’s p =0.2 and number of items (n) = 200

Hence, mean, i.e. np = 200 (0.2) =40

Q4.

Answer :

Standard deviation of the binomial distribution =4 Variance=Standard deviation2Variance, i.e. npq= 16 Mean =np = 20⇒q = 1620=45and p = 1-q = 15

Page 33.26 (Very Short Answers)

Q5.

Answer :

Mean of the binomial distribution, i.e. np =10Variance = Standard deviation2, i.e. npq =4∴q= VarianceMean =0.4

Q6.

Answer :

Given, np=4 and npq=2∴p = 1-VarianceMean = 1-24 =12and q=12 and n = npp =412 = 8Hence, the binomial distribution is given byPX=r=Cr8 12r128-r , r=0,1,2,3…….8∴P(X=1) = 8128 = 132

Q7.

Answer :
Mean =2 ,Variance =1∴ q=VarianceMean= 12and p =1-12=12n = Meanp= 212=4The binomial distribution is given byP(X=r)=Cr412r124-r∴ P(X=0)=C04120124-0 , r=0,1,2,3,4=124P(X>1) = 1-P(X=0) = 1-124 = 1516

Q8.

Answer :

In the given binomial distribution, n = 4 and
P(X=0) = 1681 Binomial distribution is given byP(X=0)=C04 p0q4-0=q4 We know that P(X=0) = 1681 ∴ q4 =1681⇒ q4=234⇒q =23

Q9.

Answer :

Mean (np)= 4

Variance (npq )= 3

.⇒q = 34 Hence, p =1-34 = 14 and n = Meanp= 4×4 =16Therefore, the binomial distribution is given byP(X=r) =Cr16 14r3416-r , r=0,1,2….rProbability of no success = C0161403416-0=3416

Q10.

Answer :

For binomial distribution of X,P(X=r)= Crn(p)r(q)n-r, r= 0,1,2,…, nP(X=1)=np(q)n-1 P(X=2) =C2np2 (q)n-2⇒np(q)n-1=C2np2 (q)n-2 =α Simplifying the above equation we get,q=n-12p ⇒2q = np-p On putting, q =1-p we get 2-2p = np-p p(n+1) =2 ….. (i)Also, P(X=1) = α⇒np(1-p)n-1 =α …..(ii)

Note: We cannot find the value of n as (i) and (ii) are not linear and hence we cannot find the value of P(X = 4)

Page 33.26 (Multiple Choice Questions)

Q1.

Answer :

(a) 9105

Let X denote the number of defective bulbs.

Hence, the binomial distribution is given by

n =5 , p = 10100=110 & q = 90100=910Hence, the distribution is given by P(X=r)=Cr5110r9105-r∴P(X=0)= 9105

Q2.

Answer :

(b) 181

In the given binomial distribution, n = 4 and
P(X=0) = 1681 Binomial distribution is given byP(X=0)=C04 p0q4-0=q4 We know that P(X=0) = 1681 ∴ q4 =1681⇒ q4=234⇒q =23∴ p=1-23=13Then , P(X=4)=C44 p4q4-4=134=181

Q3.

Answer :

(c) 7

Let p=chance of hitting a distant target
⇒p =10% or p= 0.1

⇒q =1-0.1= 0.9Let n be the least number of rounds .P(hitting atleast once) = P(X≥1) ⇒ 1-P(X=0) ≥50%⇒ 1-P(X=0) ≥ 0.5P(X=0) ≤0.5⇒(0.9)n≤0.5Taking log on both the sides, we get n log 0.9 ≤ log 0.5 ⇒n≤log 0.5log 0.9⇒n≤7.2 Therefore, 7 is the least number of rounds that he must fire in order to have more than 50% chance of hitting the target at least once.

Q4.

Answer :

(c) 15/2¹³

Let X denote the number of heads in a fixed number of tosses of a coin .Then, X is a binomial variate
with parameters n and p=12

Given that P (X=7) =P (X = 9). Also, p = q = 0.5

P(X=r) = Crn(0.5)r(0.5)n-r =Crn (0.5)n∴P(X=7) =C7n(0.5)n and P(X=9)=C9n(0.5)nIt is given that P(X=7)=P(X=9)∴ C7n(0.5)n =C9n(0.5)n⇒n!7! n-7 !=n!9! n-9 !⇒9×8=n-7n-8⇒n2-8n-7n+56=72⇒n2-15n-16=0⇒n+1n-16=0⇒n=-1 or n=16 ⇒n=- 1 (Not possible as n denotes the number of tosses of a coin)∴ n=16 Hence, P(X=2) =C216(0.5)16 = 16.152×1216= 15213

Q5.

Answer :

(a) 1/2

Here n=100
Let X denote the number of times a tail is obtained.

Here, p =q = 12P(X=odd) = P(X=1,3,5,….99) = C1100+C3100+…..+C9910012100= Sum of odd coefficients in binomial expansion in (1+x)10012100= 2(100-1)2100= 12

Q6.

Answer :

(c) 84×56610
Let p be the probabilty of obtaining a six in a single throw of the die .Then ,p =16and q =1-16=56Obtaining a fourth six in the tenth throw of the die means that in the first nine throws there are 3 sixes and the fourth six is obtained in the tenth throw. Therefore, required probability=P(Getting 3 sixes in the first nine throws) P(Getting a six in the tenth throw)=C39 p3q9-3 p=C39163566×16= 84 x 56610

Q7.

Answer :

(a) 1/2

Given that P(X=r) = k P(X=n-r), where k is independent of n and r .

Crn prqn-r = k Cn-rn pn-rqrWe have Crn=Cn-rn and also q=1-pHence, the equation changes to the following:pr(1-p)n-r = k pn-r(1-p)r⇒(1-p)n-2r = k pn-2r⇒qpn-2r=k This is possible when p =q and k becomes 1.Hence, p = q = 12

Q8.

Answer :

(a) 7, 14

Here, p=12and q=12Binomial distribution is given by P(X=r)=Crn12r12n-r

P (X = 4), P (X = 5), P(X = 6) are in A.P.

∴C4n+C6n=2C5n⇒n(n-1)(n-2)(n-3)24!+n(n-1)(n-2)(n-3)(n-4)(n-5)26!= n(n-1)(n-2)(n-3)(n-4)5!By simplifying, we get12+(n-4)(n-5)2(30)= n-45Taking LCM as 60, we get30+n2-9n+20 = 12n-48 ⇒n2-21n+98 =0 ⇒(n-7)(n-14) =0⇒n =7, 14

Page 33.27(Multiple Choice Questions)

Q9.

Answer :

(b) 51/101

Let X denote the number of coins showing head. Therefore, X follows a binomial distribution with p and n as parameters.Given that P(X=50)=P(X=51)⇒C50100p50q50 = C51100p51q49on simplifying we get,5150=pq⇒5150=p1-p (since p+q=1)⇒p=51101

Q10.

Answer :

(a) 49, 50

When a coin is tossed 99 times, the number of heads X follows a binomial distribution with
p = q=12= 0.5P(X=r) =Crn(0.5)r(0.5)n-r=Crn(0.5)nAs (0.5)n is common to all r it is enough if we find the maximum of Crn.We know that for odd number of n, there will be two equal maximum terms,i.e. when r=n-12and r = n+12Hence, n =99 So, the maximum is obtained when r = 49 or 50

Q11.

Answer :

(d) 3

Let X denote the number of coins.
Then, X follows a binomial distribution with

p = 12 , q = 12It is given that P(X≥1) ≥0.8⇒1-P(X=0) ≥0.8⇒P(X=0)≤1-0.8 ⇒P(X=0)=0.2⇒12n≤0.2 ⇒2n≥10.2⇒2n≥5This is possible when n≥3So, the least value of n is 3.

Q12.

Answer :

(d) 15/16

Mean =2 and variance =1

⇒np =2 and npq = 1⇒q=12 ⇒p = 1-12=12 n=Meanp⇒n =4Hence, the distribution is given byPX=r=Cr412r124-r , r=0,1,2,3,4∴P(X≥1) = 1-P(X=0) = 1-124=1516

Q13.

Answer :

(a) 1/3

Probability of heads = p
Probability of Tails = (1 − p)
Probability that first head appears at even turn
Pe = (1 − p)p + (1 − p)3p+(1 − p)5p + …..
= (1 − p)p (1 + (1 − p)2 + (1 − p)4+ …..)

=1-pp11-1-p2=1-pp1-p2+2p

Pe=1-p2-pPe=251-p2-p=255-5p=4-2p3p=1p=13Q14.

Q14.

Answer :

(b) 119128

n=8, p = 12∴ q=1-12=12Hence, the distribution is given byP(X=r) = Cr812r128-r PX-4≤2 = P(-2≤X-4≤2) =P(2≤X≤6)=P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)=C28+C38+C48+C58+C6828= 28+56+70+56+28256=238256=119128

Q15.

Answer :

(c) 33

The binomial distribution is given by,
fr, n, p=PX=r=crn pr 1-pn-r
This value can be maximum at a particular r, which can be determined as follows,
fr+1, n, pfr, n, p=1⇒cr+1npr+1×1-pn-r-1crnpr×1-pn-r=n-rpr+1 1-p=1
On substituting the values of n = 100, p=13, we get
100-r×13r+1 1-13=1⇒100-r13=r+123⇒100-r=r+12⇒100-r=2r+2⇒98=3r⇒3r=98∴r=983
The integer value of r satisfies (n + 1)p − 1 ≤ m < (n + 1)p

f (r, n, p) is montonically increasing for r < m and montonically decreasing for r > m

as  983≤m<1013

∴ The integer value of r is 33.

Q16.

Answer :

(b) C27×5568
Let p be the probabilty of obtaining a six in a single throw of the die .Then ,p =16and q =1-16=56Obtaining a third six in the eighth throw of the die means that in first seven throws there are 2 sixes and the third six is obtained in the eighth throw . Therefore,required probability=P(Getting 2 sixes in the first seven throws) P(Getting six in the eighth throw)=C27 p2q7-2 p=C27162565 ×16= C27 x 5568

Q17.

Answer :

Answer: (d)

Let p= probability that a selected coupon bears number ≤9.
p=915=35
n = number of coupons drawn with replacement
X = number of coupons bearing number ≤9
Probability that the largest number on the selected coupons does not exceed 9
= probability that all the coupons bear number ≤9
= P(X=7) = C77p7q0=377
Similarly, probability that largest number on the selected coupon bears the number ≤8 will be

P(X=7) = C77p7q0=8157 (since, p will become 815)
Hence required probability will be = 377-8157
So, option (d)

Q18.

Answer :

Let number be abcde
Case 1 : e = 0
a, b, c can be filled in 9 × 9 × 9 ways
c = 0 ⇒ 9 × 8 × 1 ways and d has 9 choices
c ≠ 0 ⇒ (9 × 9 × 9 – 9 × 8 × 1) = 657
in the case d has 8 choices ⇒ 657 × 8
Total case = 9 × 8 × 1 × 9 + 657 × 8 ⇒ 5904
Case 2 : e = 5
If c = 5,
if a ≠ 5 then a, b, c can be filled in 8 × 8 × 1 = 64 ways
if a = 5 then a, b, c can be filled in 1 × 9 × 1 = 9 ways
if c ≠ 5, then first 3 digits can be filled in 729 – 64 – 9 = 656 ways
here d has 8 choices
No. of member ending in 5 and no two consecutive digits being identical ⇒ (64 + 9) × 9 + 656 × 8
⇒ 5905
Total cases ⇒ 5904 + 5905 ⇒ 11809

Q19.

Answer :

(b) 105512

Let X denote the number of heads obtained in 10 tosses of a coin . Then, X follows a binomial distribution with n =6 , p=12=qThe distribution is given by P(X=r)=Cr1012r1210-r∴ P(X=6) = C610210= 10529=105512

Q20.

Answer :

(b) C6161463410

Mean (np) = 4 and Variance (npq) = 3

∴ q=34⇒p =1-34=14and n = 16Let X denotes the number of successes in 16 trials .Then, P(X=r) =Cr1614r3416-r⇒P(X=6) = Probability (getting exactly 6 successes)= 16C61463410

Q21.

Answer :

(c) 12

p =14∴ q= 1-14=34Standard deviation =3 ⇒Variance = 32=9Or npq = 9⇒ np =9q= 363=12Therefore, the mean of the given binomial distribution is 12.

Page 33.28(Multiple Choice Questions)

Q22.

Answer :

(d) 1516

Let X denote the number of heads obtained in four tosses of a coin .
Then X follows a binomial distribution with
n =4 and p =q =12Distribution is given by P(X=r)=Cr412r124-r ∴P(X=r)=C04120124-0 P(atleast one head turns up) = P(X≥1) = 1-P(X=0) = 1-124=1516

Q23.

Answer :

n =3

P(X=1) =8 P(X=3) (Given)The distribution is given by P(X=r)=Cr3prq3-rP(X=1)=C13p1q2 and P(X=3)=C33p3q0⇒3pq2 = 8p3 ⇒8p2 = 3q2 ⇒8p2= 3(1-p)2 ⇒8p2= 3-6p+3p2⇒5p2+6p-3=0⇒p=-6±9610

Hence , it does not match any of the answer choices.

Q24.

Answer :

(b) 3

Let X be the number of heads. Then X follows a binomial distribution with

p =12, q=12Hence, the distribution is given by P(X=r) = Crn12r12n-r, r=0,1,2,3……n∴P(X≥1) =1-P(X=0) = 1-12n ≥0.8Or 2n≥10.2⇒ 2n≥5This is possible only when n ≥3.So, the least value of n must be 3.

Q25.

Answer :

(d) 12

Let X be the number of males.
p = q = 12 (given)P(X=n-1) =Cn-1npn-1q1 = 3210⇒n12n = 3210 ⇒n12n = 322212 ⇒n12n= 121212By comparing the two sides, we get n =12