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NCERT Solutions for Class 12 Maths Chapter 13 - Probability

NCERT Solutions for Class 12 Maths Chapter 13 - Probability

Edited By Komal Miglani | Updated on Jun 20, 2025 09:54 AM IST | #CBSE Class 12th
Upcoming Event
CBSE Class 12th  Exam Date : 15 Jul' 2025 - 15 Jul' 2025

Probability is not a certainty, but it is the possibility with a heartbeat. In general terms, probability is how likely something is to happen. It is like a number between 0 and 1. If it is 0, it will not happen, but if it is 1, it will definitely happen. Anything in between 0 and 1 means it might happen. In the Probability Class 12 NCERT solutions, students will learn concepts of probability distributions, random variables, Bayes' theorem, and conditional probability thoroughly. The main goal of these solutions by NCERT is to prepare students for the Class 12 board exam as well as other competitive exams.

This Story also Contains
  1. NCERT Solution for Class 12 Maths Chapter 13 Solutions: Download PDF
  2. NCERT Solutions for Class 12 Maths Chapter 13: Exercise Questions
  3. Class 12 Maths NCERT Chapter 13: Extra Question
  4. Probability Class 12 Chapter 13: Topics
  5. Probability Class 12 Solutions: Important Formulae
  6. Approach to Solve Questions of Probability Class 12
  7. What Extra Should Students Study Beyond NCERT for JEE?
  8. NCERT Solutions for Class 12 Maths: Chapter Wise
NCERT Solutions for Class 12 Maths Chapter 13 - Probability
NCERT Solutions for Class 12 Maths Chapter 13 - Probability

Every day before leaving our houses, we check for weather forecasts. Ever wondered how metrologists predict the weather? Ever thought about how these clothing brands determine what’s going to trend or on what models, and how AI and machine learning work? The solution is in the Probability chapter! Experienced Careers360 teachers abiding by the latest CBSE syllabus have curated these class 12 Maths NCERT solutions of Probability. Students can additionally refer to the NCERT Exemplar Solutions for Class 12 Maths Chapter 13 Probability for better practice and understanding of the topic.

NCERT Solution for Class 12 Maths Chapter 13 Solutions: Download PDF

Students who wish to access the Class 12 Maths Chapter 13 NCERT Solutions can click on the link below to download the complete solution in PDF.

Download PDF

NCERT Solutions for Class 12 Maths Chapter 13: Exercise Questions

Class 12 Maths chapter 13 solutions - Exercise: 13.1
Page number: 413-415
Total questions: 17

Question 1: Given that E and F are events such that P(E)=0.6,P(F)=0.3 and p(EF)=0.2, find P(EF) and P(FE)

Answer:

It is given that P(E)=0.6,P(F)=0.3 and p(EF)=0.2,

P(E|F)=p(EF)P(F)=0.20.3=23

P(F|E)=p(EF)P(E)=0.20.6=13

Question 2: Compute P(AB), if P(B)=0.5 and P(AB)=0.32

Answer:

It is given that P(B)=0.5 and P(AB)=0.32

P(A|B)=p(AB)P(B)=0.320.5=0.64

Question 3: If P(A)=0.8,P(B)=0.5 and P(BA)=0.4, find

(i) P(AB)

Answer:

It is given that P(A)=0.8, P(B)=0.5 and P(B|A)=0.4

P(B|A)=p(AB)P(A)

0.4=p(AB)0.8

p(AB)=0.4×0.8

p(AB)=0.32

Question 3: If P(A)=0.8,P(B)=0.5 and P(BA)=0.4, find

(ii) P(AB)

Answer:

It is given that P(A)=0.8,P(B)=0.5 and P(BA)=0.4,

P(AB)=0.32

P(A|B)=p(AB)P(B)

P(A|B)=0.320.5

P(A|B)=3250=0.64

Question 3: If P(A)=0.8,P(B)=0.5 and P(BA)=0.4, find

(iii) P(AB)

Answer:

It is given that P(A)=0.8,P(B)=0.5

P(AB)=0.32

P(AB)=P(A)+P(B)P(AB)

P(AB)=0.8+0.50.32

P(AB)=1.30.32

P(AB)=0.98

Question 4: Evaluate P(AB), if 2P(A)=P(B)=513 and P(AB)=25

Answer:

Given in the question 2P(A)=P(B)=513 and P(AB)=25

We know that:

P(A|B)=p(AB)P(B)

25=p(AB)513

2×55×13=p(AB)

p(AB)=213

Use, p(AB)=p(A)+p(B)p(AB)

p(AB)=526+513213

p(AB)=1126

Question 5: If P(A)=611,P(B)=511 and P(AB)=711. , find

(i) P(AB)

Answer:

Given in the question

P(A)=611,P(B)=511 and P(AB)=711.

By using the formula:

p(AB)=p(A)+p(B)p(AB)

711=611+511p(AB)

p(AB)=1111711

p(AB)=411

Question 5: If P(A)=611,P(B)=511 and P(AB)=711, find

(ii) P(AB)

Answer:

It is given that - P(A)=611,P(B)=511

p(AB)=411

We know that:

P(A|B)=p(AB)P(B)

P(A|B)=411511

P(A|B)=45

Question 5: If P(A)=611,P(B)=511 and P(AB)=711, find

(iii) P(BA)

Answer:

Given in the question-

P(A)=611,P(B)=511 and p(AB)=411

Use formula

P(B|A)=p(AB)P(A)

P(B|A)=411611

P(B|A)=46=23

Question 6: A coin is tossed three times, where

(i)E : head on third toss ,F : heads on first two tosses

Answer:

The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =23=8

According to the question

E: head on third toss, F: heads on first two tosses

E={HHH,TTH,HTH,THH}

F={HHH,HHT}

EF=HHH

P(F)=28=14

P(EF)=18

P(E|F)=P(EF)P(F)

P(E|F)=1814

P(E|F)=48=12

Question 6: A coin is tossed three times, where

(ii)E : at least two heads ,F : at most two heads

Answer:

The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =23=8

According to question

E : at least two heads , F : at most two heads

E={HHH,HTH,THH,HHT}=4

F={HTH,HHT,THH,TTT,HTT,TTH,THT}=7

EF={HTH,THH,HHT}=3

P(F)=78

P(EF)=38

P(E|F)=P(EF)P(F)

P(E|F)=3878

P(E|F)=37

Question 6: A coin is tossed three times, where

(iii)E : at most two tails ,F : at least one tail

Answer:

The sample space S when a coin is tossed three times is

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}

It can be seen that the sample space (S) has 8 elements.

Total number of outcomes =23=8

According to the question

E: at most two tails, F: at least one tail

E={HHH,TTH,HTH,THH,THT,HTT,HHT}=7

F={TTT,TTH,HTH,THH,THT,HTT,HHT}=7

EF={TTH,HTH,THH,THT,HTT,HHT}=6

P(F)=78

P(EF)=68=34

P(E|F)=P(EF)P(F)

P(E|F)=3478

P(E|F)=67

Question 7: Two coins are tossed once, where

(i) E : tail appears on one coin, F : one coin shows head

Answer:

E : tail appears on one coin, F : one coin shows head

Total outcomes =4

E={HT,TH}=2

F={HT,TH}=2

EF={HT,TH}=2

P(F)=24=12

P(EF)=24=12

P(E|F)=P(EF)P(F)

P(E|F)=1212

P(E|F)=1

Question 7: Two coins are tossed once, where

(ii)E : no tail appears,F : no head appears

Answer:

E : no tail appears, F : no head appears

Total outcomes =4

E=HHF=TT

EF=ϕ

n(EF)=0

P(F)=1

P(EF)=04=0

P(E|F)=P(EF)P(F)

P(E|F)=01=0

Question 8: A die is thrown three times,

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Answer:

E : 4 appears on the third toss, F : 6 and 5 appears respectively on first two tosses

Total outcomes =63=216

E={114,124,134,144,154,164,214,224,234,244,254,264,314,
324,334,344,354,364,414,424,434,454,464,514,
524,534,544,554,564,614,624,634,644,654,664}
n(E)=36

F={651,652,653,654,655,656}

n(F)=6

EF={654}

n(EF)=1

P(EF)=1216

P(F)=6216=136

P(E|F)=P(EF)P(F)

P(E|F)=1216136

P(E|F)=16

Question 9: Mother, father and son line up at random for a family picture

E : son on one end, F : father in middle

Answer:

E : son on one end, F : father in middle

Total outcomes =3!=3×2=6

Let S be son, M be mother and F be father.

Then we have,

E={SMF,SFM,FMS,MFS}

n(E)=4

F={SFM,MFS}

n(F)=2

EF={SFM,MFS}

n(EF)=2

P(F)=26=13

P(EF)=26=13

P(E|F)=P(EF)P(F)

P(E|F)=1313

P(E|F)=1

Question 10: A black and a red dice are rolled.

(a) Find the conditional probability of obtaining a sum greater than 9 , given that the black die resulted in a 5.

Answer:

A black and a red dice are rolled.

Total outcomes =62=36

Let the A be event obtaining a sum greater than 9 and B be a event that the black die resulted in a 5.

A={46,55,56,64,65,66}

n(A)=6

B={51,52,53,54,55,56}

n(B)=6

AB={55,56}

n(AB)=2

P(AB)=236

P(B)=636

P(A|B)=P(AB)P(B)

P(A|B)=236636=26=13

Question 10: A black and a red dice are rolled.

(b) Find the conditional probability of obtaining the sum 8 , given that the red die resulted in a number less than 4 .

Answer:

A black and a red dice are rolled.

Total outcomes =62=36

Let A be the event obtaining a sum 8 and B be a event that the red die resulted in a number less than 4.

A={26,35,53,44,62,}

n(A)=5

Red dice is rolled after black dice.

B={11,12,13,21,22,23,31,32,33,41,42,43,51,52,53,61,62,63}

n(B)=18

AB={53,62}

n(AB)=2

P(AB)=236

P(B)=1836

P(A|B)=P(AB)P(B)

P(A|B)=2361836=218=19

Question 11: A fair die is rolled. Consider events E={1,3,5},F{2,3} and G={2,3,4,5} Find

(i) P(EF) and P(FE)

Answer:

A fair die is rolled.

Total oucomes ={1,2,3,4,5,6}=6

E={1,3,5},F{2,3}

EF={3}

n(EF)=1

n(F)=2

n(E)=3

P(E)=36 P(F)=26 and P(EF)=16

P(E|F)=P(EF)P(F)

P(E|F)=1626

P(E|F)=12

P(F|E)=P(FE)P(E)

P(F|E)=1636

P(F|E)=13

Question 11: A fair die is rolled. Consider events E={1,3,5},F={2,3} and G={2,3,4,5} Find

(ii) P(EG) and P(GE)

Answer:

A fair die is rolled.

Total outcomes ={1,2,3,4,5,6}=6

E={1,3,5} , G={2,3,4,5}

EG={3,5}

n(EG)=2

n(G)=4

n(E)=3

P(E)=36 P(G)=46 P(EF)=26

P(E|G)=P(EG)P(G)

P(E|G)=2646

P(E|G)=24=12

P(G|E)=P(GE)P(E)

P(G|E)=2636

P(G|E)=23

Question 11: A fair die is rolled. Consider events E={1,3,5},F={2,3} and G={2,3,4,5} Find

(iii) P((EF)G) and P((EF)G)

Answer:

A fair die is rolled.

Total outcomes ={1,2,3,4,5,6}=6

E={1,3,5},F{2,3} and G={2,3,4,5}

EG={3,5} , FG={2,3}

(EG)G={3}

P[(EG)G]=16 P(EG)=26 P(FG)=26

P((EF)|G)=P(E|G)+P(F|G)P[(EF)|G]

=P(EG)P(G)+P(FG)P(G)P((EF)G)P(G)

=2646+26461646

=24+2414

=34

P((EF)|G)=P((EF)G)P(G)

P((EF)|G)=1646

P((EF)|G)=14

Question 12: Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(i) the youngest is a girl,

Answer:

Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by G1andG2 respectively also first and second boy are denoted by B1andB2

If a family has two children, then total outcomes =22=4 ={(B1B2),(G1G2),(G1B2),(G2B1)}

Let A= both are girls ={(G1G2)}

and B= the youngest is a girl = ={(G1G2),(B1G2)}

AB={(G1G2)}

P(AB)=14 P(B)=24

P(A|B)=P(AB)P(B)

P(A|B)=1424

P(A|B)=12

Therefore, the required probability is 12.

Question 12: Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that

(ii) at least one is a girl?

Answer:

Assume that each born child is equally likely to be a boy or a girl.

Let first and second girl are denoted by G1andG2 respectively also first and second boy are denoted by B1andB2

If a family has two children, then total outcomes =22=4 ={(B1B2),(G1G2),(G1B2),(G2B1)}

Let A= both are girls ={(G1G2)}

and C= at least one is a girl = ={(G1G2),(B1G2),(G1B2)}

AB={(G1G2)}

P(AB)=14 P(C)=34

P(A|C)=P(AC)P(C)

P(A|C)=1434

P(A|C)=13

Question 13: An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?

Answer:

An instructor has a question bank consisting of 300 easy True / False questions, 200 difficult True / False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions.

Total number of questions =300+200+500+400=1400

Let A = question be easy.

n(A)=300+500=800

P(A)=8001400=814

Let B = multiple-choice question

n(B)=500+400=900

P(B)=9001400=914

AB= easy multiple questions

n(AB)=500

P(AB)=5001400=514

P(A|B)=P(AB)P(B)

P(A|B)=514914

P(A|B)=59

Therefore, the required probability is 59.

Question 14: Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.

Answer:

Two dice are thrown.

Total outcomes =62=36

Let A be the event ‘the sum of numbers on the dice is 4.

A={(13),(22),(31)}

Let B be the event that two numbers appearing on throwing two dice are different.

B={(12),(13),(14),(15),(16),(21)(23),(24),(25),(26,)(31),(32),(34),(35),(36),(41),(42),(43),(45),(46),(51),(52),(53),(54),(56),(61),(62),(63),(64),(65)} n(B)=30

P(B)=3036

AB={(13),(31)}

n(AB)=2

P(AB)=236

P(A|B)=P(AB)P(B)

P(A|B)=2363036

P(A|B)=230=115

Therefore, the required probability is 115.

Question 15: Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’, given that ‘at least one die shows a 3’.

Answer:

Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin.

Total outcomes
={(1H),(1T),(2H),(2T),(31),(32),(33),(34),(35),(36),(4H),(4T),(5H),(5T),(61),(62),(63),(64),(65),(66)}

Total number of outcomes =20

Let A be a event when coin shows a tail.

A={((1T),(2T),(4T),(5T)}

Let B be a event that at least one die shows a 3’.

B={(31),(32),(33),(34),(35),(36),(63)}

n(B)=7

P(B)=720

AB=ϕ

n(AB)=0

P(AB)=020=0

P(A|B)=P(AB)P(B)

P(A|B)=0720

P(A|B)=0

Question 16: In the following Exercise 16 choose the correct answer:

If P(A)=12,P(B)=0, then P(AB) is

(A) 0

(B) 12

(C) notdefined

(D) 1

Answer:

It is given that

P(A)=12,P(B)=0,

P(A|B)=P(AB)P(B)

P(A|B)=P(AB)0

Hence, P(A|B) is not defined.

Thus, the correct option is C.

Question 17: In the following Exercise 17 choose the correct answer:

If A and B are events such that P(AB)=P(BA), then

(A) AB but AB

(B) A=B

(C) AB=ψ

(D) P(A)=P(B)

Answer:

It is given that P(AB)=P(BA),

P(AB)P(B) =P(AB)P(A)

P(A)=P(B)

Hence, option D is correct.

Class 12 Maths chapter 13 solutions - Exercise: 13.2
Page number: 421-423
Total questions: 18

Question 1: If P(A)=35 and P(B)=15, find P(AB) if A and B are independent events.

Answer:

P(A)=35 and P(B)=15,

Given: A and B are independent events.

So we have, P(AB)=P(A).P(B)

P(AB)=35×15

P(AB)=325

Question 2: Two cards are drawn at random and without replacement from a pack of 52 playing cards. Find the probability that both the cards are black.

Answer:

Two cards are drawn at random and without replacement from a pack of 52 playing cards.

There are 26 black cards in a pack of 52.

Let P(A) be the probability that the first card is black.

Then, we have

P(A)=2652=12

Let P(B) be the probability that the second card is black.

Then, we have

P(B)=2551

The probability that both the cards are black =P(A).P(B)

=12×2551

=25102

Question 3: A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale, otherwise, it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.

Answer:

Total oranges = 15

Good oranges = 12

Bad oranges = 3

Let P(A) be the probability that first orange is good.

The, we have

P(A)=1215=45

Let P(B) be the probability that the second orange is good.

P(B)=1114

Let P(C) be the probability that the third orange is good.

P(C)=1013

The probability that a box will be approved for sale =P(A).P(B).P(C)

=45.1114.1013

=4491

Question 4: A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not.

Answer:

A fair coin and an unbiased die are tossed, then the total outputs are:

={(H1),(H2),(H3),(H4),(H5),(H6),(T1),(T2),(T3),(T4),(T5),(T6)}

=12

A is the event ‘the head appears on the coin’.

Total outcomes of A are : ={(H1),(H2),(H3),(H4),(H5),(H6)}

P(A)=612=12

B is the event ‘3 on the die’.

Total outcomes of B are : ={(T3),(H3)}

P(B)=212=16

AB=(H3)

P(AB)=112

Also, P(AB)=P(A).P(B)

P(AB)=12×16=112

Hence, A and B are independent events.

Question 5: A die marked 1,2,3 in red and 4,5,6 in green is tossed. Let A be the event, ‘the number is even,’ and B be the event, ‘the number is red’. Are A and B independent?

Answer:

Total outcomes ={1,2,3,4,5,6}=6 .

A is the event, ‘the number is even,’

Outcomes of A ={2,4,6}

n(A)=3.

P(A)=36=12

B is the event, ‘the number is red’.

Outcomes of B ={1,2,3}

n(B)=3.

P(B)=36=12

(AB)={2}

n(AB)=1

P(AB)=16

Also,

P(AB)=P(A).P(B)

P(AB)=12×12=1416

Thus, both events A and B are not independent.

Question 6: Let E and F be events with P(E)=35,P(F)=310 and P(EF)=15. Are E and F independent?

Answer:

Given :

P(E)=35,P(F)=310 and P(EF)=15.

For events E and F to be independent, we need

P(EF)=P(E).P(F)

P(EF)=35×310=95015

Hence, E and F are not independent events.

Question 7: Given that the events A and B are such that P(A)=12,P(AB)=35 and P(B)=p. Find p if they are

(i) mutually exclusive

Answer:

Given,

P(A)=12,P(AB)=35

Also, A and B are mutually exclusive means AB=ϕ .

P(AB)=P(A)+P(B)P(AB)

35=12+P(B)0

P(B)=3512=110

Question 7: Given that the events A and B are such that P(A)=12,P(AB)=35 and P(B)=p. Find p if they are

(ii) independent

Answer:

Given,

P(A)=12,P(AB)=35

Also, A and B are independent events means

P(AB)=P(A).P(B) . Also P(B)=p.

P(AB)=P(A).P(B)=p2

P(AB)=P(A)+P(B)P(AB)

35=12+pp2

p2=3512=110

p=210=15

Question 8: Let A and B be independent events with P(A)=0.3 and P(B)=0.4
Find

(i) P(AB)

Answer:

P(A)=0.3 and P(B)=0.4

Given: A and B are independent events

So, we have

P(AB)=P(A).P(B)

P(AB)=0.3×0.4=0.12

Question 8: Let A and B be independent events with P(A)=0.3 and P(B)=0.4 Find

(ii) P(AB)

Answer:

P(A)=0.3 and P(B)=0.4

Given: A and B are independent events

So, we have

P(AB)=P(A).P(B)

P(AB)=0.3×0.4=0.12

We have, P(AB)=P(A)+P(B)P(AB)

P(AB)=0.3+0.40.12=0.58

Question 8: Let A and B be independent events with P(A)=0.3 and P(B)=0.4 Find

(iii) P(AB)

Answer:

P(A)=0.3 and P(B)=0.4

Given: A and B are independent events

So, we have P(AB)=0.12

P(A|B)=P(AB)P(B)

P(A|B)=0.120.4=0.3

Question 8: Let A and B be independent events with P(A)=0.3 and P(B)=0.4 Find

(iv) P(BA)

Answer:

P(A)=0.3 and P(B)=0.4

Given: A and B are independent events

So, we have P(AB)=0.12

P(B|A)=P(AB)P(A)

P(B|A)=0.120.3=0.4

Question 9: If A and B are two events such that P(A)=14,P(B)=12 and P(AB)=18, find P(notAandnotB).

Answer:

If A and B are two events such that P(A)=14,P(B)=12 and P(AB)=18,

P(notAandnotB)=P(AB)

P(notAandnotB)=P(AB) use, (P(AB)=P(AB))

=1(P(A)+P(B)P(AB))

=1(14+1218)

=1(6818)

=158

=38

Question 10: Events A and B are such that P(A)=12,P(B)=712 and P(notAornotB)=14. State whether A and B are independent ?

Answer:

If A and B are two events such that P(A)=12,P(B)=712 and P(notAornotB)=14.

P(AB)=14

P(AB)=14 (AB=(AB))

1P(AB)=14

P(AB)=114=34

AlsoP(AB)=P(A).P(B)

P(AB)=12×712=724

As we can see 34724

Hence, A and B are not independent.

Question 11: Given two independent events A and B such that P(A)=0.3,P(B)=0.6, Find

(i) P(AandB)

Answer:

P(A)=0.3,P(B)=0.6,

Given two independent events A and B.

P(AB)=P(A).P(B)

P(AB)=0.3×0.6=0.18

Also, we know P(AandB)=P(AB)=0.18

Question 11: Given two independent events A and B such that P(A)=0.3,P(B)=0.6, Find

(ii) P(AandnotB)

Answer:

P(A)=0.3,P(B)=0.6,

Given two independent events A and B.

P(AandnotB) =P(A)P(AB)

=0.30.18=0.12

Question 11: Given two independent events A and B such that P(A)=0.3,P(B)0.6, Find

(iii) P(AorB)

Answer:

P(A)=0.3,P(B)=0.6,

P(AB)=0.18

P(AorB)=P(AB)

P(AB)=P(A)+P(B)P(AB)

=0.3+0.60.18

=0.90.18

=0.72

Question 11: Given two independent events A and B such that P(A)=0.3,P(B)=0.6, Find

(iv) P(neitherAnorB)

Answer:

P(A)=0.3,P(B)=0.6,

P(AB)=0.18

P(AorB)=P(AB)

P(AB)=P(A)+P(B)P(AB)

=0.3+0.60.18

=0.90.18

=0.72

P(neitherAnorB) =P(AB)

=P((AB))

=1P(AB)

=10.72

=0.28

Question 12: A die is tossed thrice. Find the probability of getting an odd number at least once.

Answer:

A die is tossed thrice.

Outcomes ={1,2,3,4,5,6}

Odd numbers ={1,3,5}

The probability of getting an odd number on the first throw

=36=12

The probability of getting an even number

=36=12

Probability of getting even number three times

=12×12×12=18

The probability of getting an odd number at least once
= 1 - the probability of getting an odd number in none of the throws

= 1 - probability of getting an even number three times

=118

=78

Question 13: Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(i) both balls are red.

Answer:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a red ball in the first draw

=818=49

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

=818=49

The probability that both balls are red

=49×49=1681

Question 13: Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(ii) first ball is black and second is red.

Answer:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

The probability of getting a black ball in the first draw

=1018=59

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

=818=49

The probability that the first ball is black and the second is red

=59×49=2081

Question 13: Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that

(iii) one of them is black and other is red.

Answer:

Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls.

Total balls =18

Black balls = 10

Red balls = 8

Let the first ball is black and the second ball is red.

The probability of getting a black ball in the first draw

=1018=59

The ball is replaced after drawing the first ball.

The probability of getting a red ball in the second draw

=818=49

The probability that the first ball is black and the second is red

=59×49=2081 ...........................1

Let the first ball is red and the second ball is black.

The probability of getting a red ball in the first draw

=818=49

The probability of getting a black ball in the second draw

=1018=59

the probability that the first ball is red and the second is black

=49×59=2081 ...........................2

Thus,

The probability that one of them is black and the other is red = the probability that the first ball is black and the second is red + the probability that the first ball is red and the second is black =2081+2081=4081

Question 14: Probability of solving specific problem independently by A and B are 12 and 13 respectively. If both try to solve the problem independently, find the probability that

(i) the problem is solved

Answer:

P(A)=12 and P(B)=13

Since the problem is solved independently by A and B,

P(AB)=P(A).P(B)

P(AB)=12×13

P(AB)=16

probability that the problem is solved =P(AB)

P(AB)=P(A)+P(B)P(AB)

P(AB)=12+1316

P(AB)=5616

P(AB)=46=23

Question 14: Probability of solving specific problem independently by A and B are 12 and 13, respectively. If both try to solve the problem independently, find the probability that

(ii) exactly one of them solves the problem

Answer:

P(A)=12 and P(B)=13

P(A)=1P(A) , P(B)=1P(B)

P(A)=112=12 , P(B)=113=23

probability that exactly one of them solves the problem =P(AB)+P(AB)

probability that exactly one of them solves the problem =P(A).P(B)+P(A)P(B)

=12×23+12×13

=26+16

=36=12

Question 15: One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(i) E : ‘the card drawn is a spade’

F : ‘the card drawn is an ace’

Answer:

One card is drawn at random from a well-shuffled deck of 52 cards

Total ace = 4

total spades =13

E : ‘the card drawn is a spade

F : ‘the card drawn is an ace’

P(E)=1352=14

P(F)=452=113

EF: a card which is spade and ace = 1

P(EF)=152

P(E).P(F)=14×113=152

P(EF)=P(E).P(F)=152

Hence, E and F are independent events.

Question 15: One card is drawn at random from a well-shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(ii) E : ‘the card drawn is

F : ‘the card drawn is a king’

Answer:

One card is drawn at random from a well-shuffled deck of 52 cards

Total black card = 26

total king =4

E : ‘the card drawn is black’

F : ‘the card drawn is a king’

P(E)=2652=12

P(F)=452=113

EF: a card which is black and king = 2

P(EF)=252=126

P(E).P(F)=12×113=126

P(EF)=P(E).P(F)=126

Hence, E and F are independent events.

Question 15: One card is drawn at random from a well-shuffled deck of 52 cards. In which of the following cases are the events E and F independent?

(iii) E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

Answer:

One card is drawn at random from a well-shuffled deck of 52 cards

Total king or queen = 8

total queen or jack = 8

E : ‘the card drawn is a king or queen’

F : ‘the card drawn is a queen or jack’.

P(E)=852=213

P(F)=852=213

EF: a card which is queen = 4

P(EF)=452=113

P(E).P(F)=213×213=4169

P(EF)P(E).P(F)

Hence, E and F are not independent events

Question 16: In a hostel, 60% of the students read Hindi newspapers, 40% read English newspapers and 20% read both Hindi and English newspapers. A student is selected at random.

(a) Find the probability that she reads neither Hindi nor English newspapers

Answer:

H : 60% of the students read Hindi newspaper,

E : 40% read English newspaper and

HE: 20% read both Hindi and English newspapers.

P(H)=60100=610=35

P(E)=40100=410=25

P(HE)=20100=210=15

The probability that she reads neither Hindi nor English newspapers =1P(HE)

=1(P(H)+P(E)P(HE))

=1(35+2515)

=145

=15

Question 16: In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

(b) If she reads Hindi newspaper, find the probability that she reads English newspaper.

Answer:

H : 60% of the students read a Hindi newspaper,

E : 40% read an English newspaper and

HE: 20% read both Hindi and English newspapers.

P(H)=60100=610=35

P(E)=40100=410=25

P(HE)=20100=210=15

The probability that she reads English newspapers if she reads Hindi newspapers =P(E|H)

P(E|H)=P(EH)P(H)

P(E|H)=1535

P(E|H)=13

Question 16: In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random.

(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.

Answer:

H : 60% of the students read Hindi newspapers,

E : 40% read English newspaper and

HE: 20% read both Hindi and English newspapers.

P(H)=60100=610=35

P(E)=40100=410=25

P(HE)=20100=210=15

The probability that she reads a Hindi newspaper if she reads an English newspaper =P(H|E)

P(H|E)=P(HE)P(E)

P(H|E)=1525

P(H|E)=12

Question 17: The probability of obtaining an even prime number on each die, when a pair of dice is rolled is

(A) 0

(B) 13

(C) 112

(D) 136

Answer:

When a pair of dice is rolled, total outcomes =62=36

Even prime number ={2}

n(evenprimenumber)=1

The probability of obtaining an even prime number on each die =P(E)

P(E)=136

Option D is correct.

Question 18: Two events A and B will be independent if

(A) A and B are mutually exclusive

(B) P(AB)=[1P(A)][1P(B)]

(C) P(A)=P(B)

(D) P(A)+P(B)=1

Answer:

Two events A and B will be independent if

P(AB)=P(A).P(B)

Or P(AB)=P(AB)=P(A).P(B)=(1P(A)).(1P(B))

Option B is correct.

Class 12 Maths chapter 13 solutions - Exercise: 13.3
Page number: 431-433
Total questions: 14

Question 1: An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red?

Answer:

Black balls = 5

Red balls = 5

Total balls = 10

CASE 1 Let red ball be drawn in first attempt.

P(drawingredball)=510=12

Now two red balls are added in urn .

Now red balls = 7, black balls = 5

Total balls = 12

P(drawingredball)=712

CASE 2

Let black ball be drawn in first attempt.

P(drawingblackball)=510=12

Now two black balls are added in urn .

Now red balls = 5, black balls = 7

Total balls = 12

P(drawingredball)=512

The probability that the second ball is red =

=12×712+12×512

=724+524

=1224=12

Question 2: A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.

Answer:

BAG 1 : Red balls =4 Black balls=4 Total balls = 8

BAG 2 : Red balls = 2 Black balls = 6 Total balls = 8

B1 : selecting bag 1

B2 : selecting bag 2

P(B1)=P(B2)=12

Let R be an event of getting red ball

P(R|B1)=P(drawingredballfromfirstbag)=48=12

P(R|B2)=P(drawingredballfromsecondbag)=28=14

Probability that the ball is drawn from the first bag,

given that it is red is P(B1|R) .

Using Bayes' theorem, we have

P(B1|R)=P(B1).P(R|B1)P(B1).P(R|B1)+P(B2).P(R|B2)

P(B1|R)=12×1212×12+12×14

P(B1|R)=1414+18

P(B1|R)=1438

P(B1|R)=23

Question 3: Of the students in a college, it is known that 60% reside in hostel and 40% are day scholars (not residing in hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student is chosen at random from the college and he has an A grade, what is the probability that the student is a hostlier?

Answer:

H : resides in a hostel

D : day scholars

A : students who attain grade A

P(H)=60100=610=35

P(D)=40100=410=25

P(A|H)=30100=310

P(A|D)=20100=210=15

By Bayes' theorem :

P(H|A)=P(H).P(A|H)P(H).P(A|H)+P(D).P(A|D)

P(H|A)=35×31035×310+25×15

P(H|A)=950950+225

P(H|A)=9501350

P(H|A)=913

Question 4: In answering a question on a multiple choice test, a student either knows the answer or guesses. Let 34 be the probability that he knows the answer and 14 be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability 14 . What is the probability that the student knows the answer given that he answered it correctly?

Answer:

A : Student knows the answer.

B : Student, guess the answer

C : Answer is correct

P(A)=34 P(B)=14

P(C|A)=1

P(C|B)=14

By Bayes' theorem :

P(A|C)=P(A).P(C|A)P(A).P(C|A)+P(B).P(C|B)

P(A|C)=34×134×1+14×14

=3434+116 =341316

P(A|C)=1213

Question 5: A laboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1% of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?

Answer:

A : Person selected is having the disease

B : Person selected is not having the disease.

C :Blood result is positive.

P(A)=0.1%=11000=0.001

P(B)=1P(A)=10.001=0.999

P(C|A)=99%=0.99

P(C|B)=0.5%=0.005

By Bayes' theorem :

P(A|C)=P(A).P(C|A)P(A).P(C|A)+P(B).P(C|B)

=0.001×0.990.001×0.99+0.999×0.005

=0.000990.00099+0.004995

=0.000990.005985 =9905985

=22133

Question 6: There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

Answer:

Given : A : chossing a two headed coin

B : chossing a biased coin

C : chossing a unbiased coin

P(A)=P(B)=P(C)=13

D : event that coin tossed shows a head.

P(D|A)=1

Biased coin that comes up heads 75% of the time.

P(D|B)=75100=34

P(D|C)=12

P(B|D)=P(B).P(D|B)P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)

P(B|D)=13×113×1+13×34+13×12

P(B|D)=1313+14+16

P(B|D)=13912

P(B|D)=1×123×9

P(B|D)=49

Question 7: An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accidents are 0.01,0.03 , and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Answer:

Let A : scooter drivers = 2000

B : car drivers = 4000

C : truck drivers = 6000

Total drivers = 12000

P(A)=200012000=16=0.16

P(B)=400012000=13=0.33

P(C)=600012000=12=0.5

D : the event that a person meets with an accident.

P(D|A)=0.01

P(D|B)=0.03

P(D|C)=0.15

P(A|D)=P(A).P(D|A)P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)

P(A|D)=0.16×0.010.16×0.01+0.33×0.03+0.5×0.15

P(A|D)=0.00160.0016+0.0099+0.075

P(A|D)=0.00160.0865

P(A|D)=0.019

Question 8: A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

Answer:

A : Items produced by machine A =60%

B : Items produced by machine B =40%

P(A)=60100=35

P(B)=40100=25

X : Produced item found to be defective.

P(X|A)=2100=150

P(X|B)=1100

P(B|X)=P(B).P(X|B)P(B).P(X|B)+P(A).P(X|A)

P(B|X)=25×110025×1100+35×150

P(B|X)=12501250+3250

P(B|X)=12504250

P(B|X)=14

Hence, the probability that a defective item was produced by machine B =

P(B|X)=14 .

Question 9: Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.

Answer:

A: the first groups will win

B: the second groups will win

P(A)=0.6

P(B)=0.4

X: Event of introducing a new product.

Probability of introducing a new product if the first group wins : P(X|A)=0.7

Probability of introducing a new product if the second group wins : P(X|B)=0.3

P(B|X)=P(B).P(X|B)P(B).P(X|B)+P(A).P(X|A)

p(B|X)=0.4×0.30.4×0.3+0.6×0.7

p(B|X)=0.120.12+0.42

p(B|X)=0.120.54

p(B|X)=1254

p(B|X)=29

Hence, the probability that the new product introduced was by the second group :

p(B|X)=29

Question 10: Suppose a girl throws a die. If she gets a 5 or 6 , she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 , she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3 or 4 with the die?

Answer:

Let A: Outcome on die is 5 or 6.

B: Outcome on die is 1,2,3,4

P(A)=26=13

P(B)=46=23

X: Event of getting exactly one head.

Probability of getting exactly one head when she tosses a coin three times : P(X|A)=38

Probability of getting exactly one head when she tosses a coin one time : P(X|B)=12

P(B|X)=P(B).P(X|B)P(B).P(X|B)+P(A).P(X|A)

P(B|X)=23×1223×12+13×38

P(B|X)=1313+18

P(B|X)=131124

P(B|X)=1×243×11=811

Hence, the probability that she threw 1,2,3 or 4 with the die =

P(B|X)=811

Question 11: A manufacturer has three machine operators A,B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items, respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Answer:

Let A: time consumed by machine A =50%

B: time consumed by machine B =30%

C: time consumed by machine C =20%

Total drivers = 12000

P(A)=50100=12

P(B)=30100=310

P(C)=20100=15

D: Event of producing defective items

P(D|A)=1100

P(D|B)=5100

P(D|C)=7100

P(A|D)=P(A).P(D|A)P(B).P(D|B)+P(A).P(D|A)+P(C).P(D|C)

P(A|D)=12×110012×1100+310×5100+15×7100

P(A|D)=12×11001100(12+32+75)

P(A|D)=12(175)

P(A|D)=534

Hence, the probability that a defective item was produced by A
= P(A|D)=534

Question 12: A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Answer:

Let A : Event of choosing a diamond card.

B : Event of not choosing a diamond card.

P(A)=1352=14

P(B)=3952=34

X : The lost card.

If lost card is diamond, then 12 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 12 diamond cards in 12C2 ways.

Similarly, two cards are drawn out of 51 cards in 51C2 ways.

Probability of getting two diamond cards when one diamond is lost : P(X|A)=12C251C2

P(X|A)=12!10!×2!×49!×2!51!

P(X|A)=11×1250×51

P(X|A)=22425

If the lost card is not diamond, then 13 diamond cards are left out of 51 cards.

Two diamond cards are drawn out of 13 diamond cards in 13C2 ways.

Similarly, two cards are drawn out of 51 cards in 51C2 ways.

Probability of getting two diamond cards when one diamond is not lost : P(X|B)=13C251C2

P(X|B)=13!11!×2!×49!×2!51!

P(X|B)=13×1250×51

P(X|B)=26425

The probability of the lost card being a diamond : P(B|X)

P(B|X)=P(B).P(X|B)P(B).P(X|B)+P(A).P(X|A)

P(B|X)=14×2242514×22425+34×26425

P(B|X)=11225

P(B|X)=1150

Hence, the probability of the lost card being a diamond :

P(B|X)=1150

Question 13: Probability that A speaks truth is 45 . A coin is tossed. A reports that a head appears. The probability that actually there was head is

(A) 45

(B) 12

C) 15

(D) 25

Answer:

Let A : A speaks truth

B : A speaks false

P(A)=45

P(B)=145=15

X : Event that head appears.

A coin is tossed, outcomes are head or tail.

Probability of getting a head whether A speaks thruth or not is 12

P(X|A)=P(X|B)=12

P(A|X)=P(A).P(X|A)P(B).P(X|B)+P(A).P(X|A)

P(A|X)=45×1245×12+15×12

P(A|X)=4545+15

P(A|X)=4511

P(A|X)=45

The probability that actually there was head is P(A|X)=45

Hence, option A is correct.

Question 14: If A and B are two events such that AB and P(B)0, then which of the following is correct?

(A) P(AB)=P(B)P(A)

(B) P(AB)<P(A)

(C) P(AB)P(A)

(D) None of these

Answer:

If AB and P(B)0, then

(AB)=A

Also, P(A)<P(B)

P(A|B)=P(AB)P(B)=P(A)P(B)

We know that P(B)1

11P(B)

P(A)P(A)P(B)

P(A)P(A|B)

Hence, we can see that option C is correct.

Class 12 Maths chapter 13 solutions - Miscellaneous Exercise
Page number: 435-437
Total questions: 13

Question 1(i): A and B are two events such that P(A)0. Find P(BA), if

A is a subset of B

Answer:

A and B are two events such that P(A)0.

AB

AB=A

P(AB)=P(BA)=P(A)

P(B|A)=P(BA)P(A)

P(B|A)=P(A)P(A)

P(B|A)=1

Question 1(ii): A and B are two events such that P(A)0. Find P(BA), if

AB=ϕ

Answer:

A and B are two events such that P(A)0.

P(AB)=P(BA)=0

P(B|A)=P(BA)P(A)

P(B|A)=0P(A)

P(B|A)=0

Question 2 (i): A couple has two children,

Find the probability that both children are males, if it is known that at least one of the children is male.

Answer:

A couple has two children,

sample space ={(b,b),(g,g),(b,g),(g,b)}

Let A be both children are males and B is at least one of the children is male.

(AB)={(b,b)}

P(AB)=14

P(A)=14

P(B)=34

P(A|B)=P(AB)P(B)

P(A|B)=1434=13

Question 2 (ii): A couple has two children,

Find the probability that both children are females, if it is known that the elder child is a female.

Answer:

A couple has two children,

sample space ={(b,b),(g,g),(b,g),(g,b)}

Let A be both children are females and B be the elder child is a female.

(AB)={(g,g)}

P(AB)=14

P(A)=14

P(B)=24

P(A|B)=P(AB)P(B)

P(A|B)=1424=12

Question 3: Suppose that 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

Answer:

We have 5% of men and 0.25% of women have grey hair.

Percentage of people with grey hairs =(5+0.25)%=5.25%

The probability that the selected haired person is male :

=55.25=2021

Question 4: Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

Answer:

90% of people are right-handed.

P(righthanded)=910

P(lefthanded)=q=1910=110

at most 6 of a random sample of 10 people are right-handed.

the probability that more than 6 of a random sample of 10 people are right-handed is given by,

T1010CrPrq10r=T1010Cr910r(110)10r

The probability that at most 6 of a random sample of 10 people are right-handed is given by

=1T1010Cr910r(11010r)

Question 5: If a leap year is selected at random, what is the chance that it will contain 53 Tuesdays?

Answer:

In a leap year, there are 366 days.

In 52 weeks, there are 52 Tuesdays.

The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesdays.

The remaining 2 days can be :

1. Monday and Tuesday

2. Tuesday and Wednesday

3. Wednesday and Thursday

4. Thursday and Friday

5. Friday and Saturday

6. Saturday and Sunday

7. Sunday and Monday

Total cases = 7.

Favorable cases = 2

Probability of having 53 Tuesdays in a leap year = P.

P=27

Question 6: Suppose we have four boxes A, B, C and D containing coloured marbles as given below:

1648018018261

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn

(i) from box A?

(ii) from box B?

(iii) from box C?

Answer:

1648018246557 '

(i)
Let R be the event of drawing a red marble.

Let EA,EB,EC respectively denote the event of selecting box A, B, C.

Total marbles = 40

Red marbles =15

P(R)=1540=38

Probability of drawing red marble from box A is P(EA|R)

P(EA|R)=P(EAR)P(R)

=14038

=115

(ii)
Let R be the event of drawing a red marble.

Let EA,EB,EC respectively denote events of selecting box A, B, and C.

Total marbles = 40

Red marbles =15

P(R)=1540=38

Probability of drawing red marble from box B is P(EB|R)

P(EB|R)=P(EBR)P(R)

=64038

=25

(iii)
Let R be the event of drawing a red marble.

Let EA,EB,EC respectively denote the event of selecting box A, B, and C.

Total marbles = 40

Red marbles =15

P(R)=1540=38

Probability of drawing red marble from box C is P(EC|R)

P(EC|R)=P(ECR)P(R)

=84038

=815

Question 7: Assume that the chances of a patient having a heart attack is 40%. It is also assumed that a meditation and yoga course reduces the risk of heart attack by 30% and prescription of certain drug reduces its chances by 25%. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options, the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

Answer:

Let A, E1, and E2 respectively denote the event that a person has a heartbreak, the selected person followed the course of yoga and meditation, and the person adopted

the drug prescription.

P(A)=0.40

P(E1)=P(E2)=12

P(A|E1)=0.40×0.70=0.28

P(A|E2)=0.40×0.75=0.30

The probability that the patient followed a course of meditation and yoga is P(E1,A)

P(E1,A)=P(E1).P(E1|A)P(E1).P(E1|A)+P(E2).P(E2|A)

P(E1,A)=0.5×0.280.5×0.28+0.5×0.30

=1429

Question 8: If each element of a second-order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability 12 ).

Answer:

Total number of determinants of the second order, with each element being 0 or 1 is 24=16

The values of determinant is positive in the following cases [1001],[1101],[1011]

Probability is

=316

Question 9(i): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = 0.2

P(B fails alone) = 0.15

P(A and B fail) = 0.15

Evaluate the following probabilities

P(AfailsBhasfailed)

Answer:

Let the event in which A fails and B fails be EA,EB

P(EA)=0.2

P(EAandEB)=0.15

P(Bfailsalone)=P(EB)P(EAandEB)

0.15=P(EB)0.15

P(EB)=0.3

P(EA|EB)=P(EAEB)P(EB)

=0.150.3=0.5

Question 9(ii): An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = 0.2

P(B fails alone) = 0.15

P(A and B fail) = 0.15

Evaluate the following probabilities

P(Afailsalone)

Answer:

Let event in which A fails and B fails be EA,EB

P(EA)=0.2

P(EAandEB)=0.15

P(Bfailsalone)=P(EB)P(EAandEB)

0.15=P(EB)0.15

P(EB)=0.3

P(Afailsalone)=P(EA)P(EAandEB)

=0.20.15=0.05

Question 10: Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Answer:

Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.

P(E1)=37 and P(E2)=47

Let A be the event that ball drawn is red.

When a red ball is transfered from bag 1 to bag 2.

P(A|E1)=510=12

When a black ball is transferred from bag 1 to bag 2.

P(A|E2)=410=25

P(E2|A)=P(E2).P(A|E2)P(E2).P(A|E2)+P(E1).P(A|E1)

=47×2547×25+37×12

=1631

Question 11: If A and B are two events such that P(A0) and P(BA)=1, then

Choose the correct answer of the following:

(A) AB

(B) BA

(C) B=ϕ

(D) A=ϕ

Answer:

A and B are two events such that P(A0) and P(BA)=1,

P(B|A)=P(BA)P(A)

1=P(BA)P(A)

P(BA)=P(A)

AB

Option A is correct.

Question 12: If P(AB)>P(A) , then which of the following is correct :

(A) P(BA)<P(B)

(B) P(AB)<P(A).P(B)

(C) P(BA)>P(B)

(D) P(BA)=P(B)

Answer:

P(AB)>P(A)

P(AB)P(B)>P(A)

P(AB)>P(A).P(B)

P(AB)P(A)>P(B)

P(B|A)>P(B)

Option C is correct.

Question 13: If A and B are any two events such that P(A)+P(B)P(AandB)=P(A), then

(A) P(BA)=1

(B) P(AB)=1

(C) P(BA)=0

(D) P(AB)=0

Answer:

P(A)+P(B)P(AandB)=P(A),

P(A)+P(B)P(AB)=P(A)

P(B)P(AB)=0

P(B)=P(AB)

P(A|B)=P(AB)P(B)=P(B)P(B)=1

Option B is correct.

Also, read,

Class 12 Maths NCERT Chapter 13: Extra Question

Question: Eight coins are tossed together. The probability of getting exactly 3 heads is:

Solution:
Given:
probability distribution P(X=r)=nCr(p)rqnr
The total number of coins is tossed, n=8
The probability of getting head, p=12
The probability of getting tail, q=12
The Required probability

=8C3(12)3(12)83=8×7×63×2×128=732

Hence, the correct answer is 732.

Probability Class 12 Chapter 13: Topics

The topics discussed in the NCERT Solutions for class 12, chapter 13, Probability are:

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Probability Class 12 Solutions: Important Formulae

Conditional Probability:

Conditional probability is the likelihood of an event occurring based on the occurrence of a preceding event. For two events A and B with the same sample space, the conditional probability of event A given that B has occurred (P(A|B)) is defined as:

P ( A ∣ B ) = P ( A ∩ B ) P ( B ) (when P(B) ≠ 0)

Other conditional probability relationships:

P ( S ∣ F ) = P ( F ∣ F ) = 1 P ( ( A ∪ B ) ∣ F ) = P ( A ∣ F ) + P ( B ∣ F ) − P ( ( A ∩ B ) ∣ F ) P ( E ′ ∣ F ) = 1 − P ( E ∣ F )

Multiplication Rule: The multiplication rule relates the probability of two events E and F in a sample space S:

P ( E ∩ F ) = P ( E ) ⋅ P ( F ∣ E ) = P ( F ) ⋅ P ( E ∣ F ) (when P(E) ≠ 0 and P(F) ≠ 0)

Independent Events: Two experiments are considered independent if the probability of the events E and F occurring simultaneously is the product of their individual probabilities:

P ( E ∩ F ) = P ( E ) ⋅ P ( F )

Bayes’ Theorem:

Bayes’ theorem deals with events E1, E2, …, En that form a partition of the sample space S. It allows the calculation of the probability of event Ei given event A:

P ( E i ∣ A ) = P ( E i ) ⋅ P ( A ∣ E i ) ∑ j = 1 n P ( E j ) ⋅ P ( A ∣ E j ) , for i = 1, 2, …, n

Theorem of Total Probability:

Given a partition E1, E2, …, En of the sample space and an event A, the theorem of total probability states:

P ( A ) = P ( E 1 ) ⋅ P ( A ∣ E 1 ) + P ( E 2 ) ⋅ P ( A ∣ E 2 ) + ⋯ + P ( E n ) ⋅ P ( A ∣ E n )

Approach to Solve Questions of Probability Class 12

  • Master the conditional probability formula.
    • P(AB)=P(AB)P(B), provided P(B)0.
  • Understand the types of events, such as simple, compound, mutually exclusive, exhaustive, independent or complementary.
  • In case of reverse probability, use Bayes’ theorem. The formula looks a bit complicated, but with adequate practice, it will become simpler.
  • In case of a partition of a sample case (when an event depends on several other events), use the total probability theorem.
  • Focus on the keywords like “At least”, “only if”, “given”, and “either”.
  • Use a Venn diagram for conditional probability when the numbers are small.
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NCERT Solutions for Class 12 Maths: Chapter Wise


The links below allow students to access all the Maths solutions from the NCERT book.

Also, read,

NCERT solutions for class 12 Subject-wise

Here are the subject-wise links for the NCERT solutions of class 12:

NCERT Solutions class-wise

Given below are the class-wise solutions of class 12 NCERT:

NCERT Books and NCERT Syllabus

Here are some useful links for the NCERT books and the NCERT syllabus for class 12

Frequently Asked Questions (FAQs)

1. How to solve Bayes' Theorem questions in Class 12 Probability?

To solve Bayes' Theorem questions, apply the formula:

P(A | B) = (P(B | A) x P(A)) / P(B)
Here, P(A | B) is the probability of event A given event B, P(B | A) is the probability of event B given A, and P(A) and P(B) are the probabilities of A and B.

2. What is the formula for Conditional Probability in NCERT Class 12 Maths?

The formula for Conditional Probability is:

P(A | B) = P(A ∩ B) / P(B)

where P(A | B) is the probability of A occurring given that B has occurred.

3. What are the basic concepts of Probability in NCERT Class 12 Maths?

The basic concepts include understanding random experiments, sample spaces, events (simple and compound), probability rules (addition and multiplication rules), and probability distributions.

4. How to solve Bernoulli Trials and Binomial Distribution questions?

To solve these questions, use the Binomial Distribution formula:

P(X = k) = (n choose k) x p^k x (1 - p)^(n - k)

where n is the number of trials, k is the number of successes, and p is the probability of success in each trial.

5. What is the difference between Classical and Conditional Probability in NCERT?

Classical Probability is based on equally likely outcomes (i.e., all outcomes are equally probable), while Conditional Probability is the probability of an event occurring given that another event has already occurred. The key difference is that Classical looks at all possible outcomes equally, while Conditional focuses on a subset of outcomes based on additional information.

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Changing from the CBSE board to the Odisha CHSE in Class 12 is generally difficult and often not ideal due to differences in syllabi and examination structures. Most boards, including Odisha CHSE , do not recommend switching in the final year of schooling. It is crucial to consult both CBSE and Odisha CHSE authorities for specific policies, but making such a change earlier is advisable to prevent academic complications.

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

It's understandable to feel disheartened after facing a compartment exam, especially when you've invested significant effort. However, it's important to remember that setbacks are a part of life, and they can be opportunities for growth.

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  5. Seek Support:

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Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.







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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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