**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 E_{1}, E_{2} and E_{3} 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 E_{1}, E_{2} and E_{3} 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 E_{1}, E_{2} and E_{3} 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 E_{1} and E_{2} 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(E_{1}) = Probability that the first group wins the competition = 0.6

P(E_{2}) = Probability that the second group wins the competition = 0.4

P(A/E_{1}) = Probability of introducing a new product if the first group wins = 0.7

P(A/E_{2}) = 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 E_{1}, E_{2} and E_{3} 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 E_{1}, E_{2} and E_{3} 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 E_{1}, E_{2} and E_{3} 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 E_{1}, E_{2} and E_{3} 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(E_{1}) = 10006000=16

P(E_{2}) =20006000=13

P(E_{3}) = 30006000=12

The probability that the defective bolt is produced by machine X is given by P (E_{1}/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