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RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability

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Our RD Sharma Textbook Solutions are considered extremely helpful for solving the tough questions which are asked in the CBSE Class 12 exam. TopperLearning Textbook Solutions are compiled by subject experts. Herein, you can find all the answers to the textbook questions for Chapter 31 - Probability.

All our solutions are created in accordance with the latest CBSE syllabus, and they are amended from time to time to provide the most relevant answers. Our free RD Sharma Solutions for CBSE Class 12 Mathematics will strengthen your fundamentals in Mathematics and will help you in your attempts to score more marks in the final examination. CBSE Class 12 students can refer to our solutions any time — while doing their homework and while preparing for the exam.

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RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.1

Solution 1

Solution 2


Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.2

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6(i)

Solution 6(ii)

Solution 6(iii)

Solution 7

Solution 8

 

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.3

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5(i)

Solution 5(ii)

Solution 5(iii)

Solution 5(iv)

  

Solution 6

Solution 7

Solution 8


Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

(i) Let 'A' be the event that both the children born are girls.

Let 'B' be the event that the youngest is a girl.

We have to find conditional probability P(A/B).

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction end cell row cell straight A subset of straight B rightwards double arrow straight A intersection straight B equals straight A end cell row cell rightwards double arrow straight P left parenthesis straight A intersection straight B right parenthesis equals straight P left parenthesis straight A right parenthesis equals straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half equals 1 fourth end cell row cell straight P left parenthesis straight B right parenthesis equals straight P left parenthesis BG right parenthesis plus straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half plus 1 half cross times 1 half equals 1 fourth plus 1 fourth equals 1 half end cell row cell text Hence ,  end text straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator 1 divided by 4 over denominator 1 divided by 2 end fraction equals 1 half end cell end table end style

(ii) Let 'A' be the event that both the children born are girls.

Let 'B' be the event that at least one is a girl.

We have to find the conditional probability P(A/B).

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator straight P left parenthesis straight A intersection straight B right parenthesis over denominator straight P left parenthesis straight B right parenthesis end fraction end cell row cell straight A subset of straight B rightwards double arrow straight A intersection straight B equals straight A end cell row cell rightwards double arrow straight P left parenthesis straight A intersection straight B right parenthesis equals straight P left parenthesis straight A right parenthesis equals straight P left parenthesis GG right parenthesis equals 1 half cross times 1 half equals 1 fourth end cell row cell straight P left parenthesis straight B right parenthesis equals 1 minus straight P left parenthesis BB right parenthesis equals 1 minus 1 half cross times 1 half equals 1 minus 1 fourth equals 3 over 4 end cell row cell text Hence ,  end text straight P left parenthesis straight A divided by straight B right parenthesis equals fraction numerator 1 divided by 4 over denominator 3 divided by 4 end fraction equals 1 third end cell end table end style

 

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.4

Solution 1(i)

Solution 1(ii)

Solution 1(iii)

Solution 2

Solution 3(i)

Solution 3(ii)

Solution 3(iii)

Solution 4


Solution 5

Solution 6


Solution 7

Solution 8


Solution 9


Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19


Solution 20

Solution 21

Solution 22


Solution 23

Given that the events 'A coming in time' and 'B coming in time' are independent.

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let  ' A '  denote   the   event   of  ' A   coming   in   time '.  end text end cell row cell text Then , ' end text stack text A end text with bar on top apostrophe text   denotes   the   complementary   event   of   A. end text end cell row cell text Similarly   we   define   B   and   end text stack text B end text with bar on top. end cell end table end style

begin mathsize 12px style table attributes columnalign left end attributes row cell straight P left parenthesis only text   one   coming   in   time end text right parenthesis equals straight P left parenthesis straight A intersection straight B with bar on top right parenthesis plus straight P left parenthesis straight A with bar on top intersection straight B right parenthesis end cell row cell equals straight P left parenthesis straight A right parenthesis cross times straight P left parenthesis straight B with bar on top right parenthesis plus straight P left parenthesis straight A with bar on top right parenthesis cross times straight P left parenthesis straight B right parenthesis... left parenthesis text since   A   and   B   are   independent   events end text right parenthesis end cell row cell equals 3 over 7 cross times 2 over 7 plus 4 over 7 cross times 5 over 7 equals 6 over 49 plus 20 over 49 equals 26 over 49 end cell end table end style


The advantage of coming to school in time is that you will not miss any part of the lecture and will be able to learn more.

 

Solution 24

  

Solution 25

  

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.5

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6


Solution 7

Solution 8

Solution 9

Solution 10


Solution 11

Solution 12

Solution 13

Solution 14


Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24


Solution 25


Solution 26


Solution 27


Solution 28


Solution 29

Solution 30


Solution 31


Solution 32

Solution 33

Solution 34

Solution 35

begin mathsize 12px style table attributes columnalign left end attributes row cell text Probability   of   getting   six   in   any   toss   of   a   dice end text equals 1 over 6 end cell row cell text Probability   of   not   getting   six   in   any   toss   of   a   dice end text equals 5 over 6 end cell row cell text A   and   B   toss   the   die   alternatively .  end text end cell row cell text Hence   probability   of   A ' s   win end text end cell row cell equals straight P left parenthesis straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A right parenthesis plus....... end cell row cell equals 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus..... end cell row cell equals 1 over 6 plus open parentheses 5 over 6 close parentheses squared 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 4 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 6 1 over 6 plus..... end cell row cell equals fraction numerator 1 divided by 6 over denominator 1 minus open parentheses 5 divided by 6 close parentheses squared end fraction equals 1 over 6 cross times 36 over 11 equals 6 over 11 end cell row cell Similarly comma text   probability   of   B ' s   win end text end cell row cell equals straight P left parenthesis straight A with bar on top straight B right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B right parenthesis plus straight P left parenthesis straight A with bar on top straight B with bar on top straight A with bar on top straight B with bar on top straight A with bar on top straight B right parenthesis plus...... end cell row cell equals 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 5 over 6 cross times 1 over 6 plus..... end cell row cell equals 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses squared 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 4 5 over 6 cross times 1 over 6 plus open parentheses 5 over 6 close parentheses to the power of 6 5 over 6 cross times 1 over 6 plus..... end cell row cell equals fraction numerator 5 over 6 cross times 1 over 6 over denominator 1 minus open parentheses 5 over 6 close parentheses squared end fraction equals 5 over 36 cross times 36 over 11 equals 5 over 11 end cell row cell text Since   the   probabilities   are   not   equal , end text end cell row cell text the   decision   of   the   refree   was   not   a   fair   one. end text end cell end table end style

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.6

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5


Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

  

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31.7

Solution 1



Solution 2

Solution 3


Solution 4


Solution 5

  

Solution 6

Solution 7

Solution 8


Solution 9

Solution 10

Solution 11

Solution 12


Solution 13

Solution 14

  

Solution 15

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let   end text straight E subscript 1 comma straight E subscript 2 comma straight E subscript 3 text   be   the   events   that   we   choose   the   first   coin , end text end cell row cell text second   coin ,  and   third   coin   respectively   in   a   random   toss. end text end cell row cell straight P left parenthesis straight E subscript 1 right parenthesis equals 1 third comma straight P left parenthesis straight E subscript 2 right parenthesis equals 1 third comma straight P left parenthesis straight E subscript 3 right parenthesis equals 1 third end cell row cell text Let   A   denote   the   event   when   the   toss   shows   heads. end text end cell row cell text It   is   given   that end text end cell row cell straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis equals 1 comma straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis equals 0.75 comma straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis equals.60 end cell row cell text We   have   to   find   end text straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis. end cell row cell By text   Baye ' s   theorem end text end cell row cell straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis equals fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus straight P left parenthesis straight E subscript 2 right parenthesis straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis plus straight P left parenthesis straight E subscript 3 right parenthesis straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis end fraction equals end cell row cell equals fraction numerator 1 third left parenthesis 1 right parenthesis over denominator 1 third left parenthesis 1 right parenthesis plus 1 third left parenthesis 0.75 right parenthesis plus 1 third left parenthesis 0.60 right parenthesis end fraction equals fraction numerator 1 divided by 3 over denominator left parenthesis 1 divided by 3 right parenthesis plus left parenthesis 1 divided by 4 right parenthesis plus left parenthesis 1 divided by 5 right parenthesis end fraction end cell row cell equals fraction numerator 1 divided by 3 over denominator 47 divided by 60 end fraction equals 20 over 47 end cell end table end style

 

Solution 16

Solution 17

Solution 18


Solution 19

begin mathsize 12px style table attributes columnalign left end attributes row cell text Let   end text straight E subscript 1 comma straight E subscript 2 comma straight E subscript 3 text   be   the   events   that   the   people   are   end text end cell row cell text smokers   and   non-vegetarian ,  smokers   and   vegetarian ,  end text end cell row cell text and   non-smokers   and   vegetarian   respectively. end text end cell row cell straight P left parenthesis straight E subscript 1 right parenthesis equals 2 over 5 comma straight P left parenthesis straight E subscript 2 right parenthesis equals 1 fourth comma straight P left parenthesis straight E subscript 3 right parenthesis equals 7 over 20 end cell row cell text Let   A   denote   the   event   that   the   person   has   the   special   chest   disease. end text end cell row cell text It   is   given   that end text end cell row cell straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis equals 0.35 comma straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis equals 0.20 comma straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis equals 0.10 end cell row cell text We   have   to   find   end text straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis. end cell row cell By text   Baye ' s   theorem end text end cell row cell straight P left parenthesis straight E subscript 1 divided by straight A right parenthesis equals fraction numerator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis over denominator straight P left parenthesis straight E subscript 1 right parenthesis straight P left parenthesis straight A divided by straight E subscript 1 right parenthesis plus straight P left parenthesis straight E subscript 2 right parenthesis straight P left parenthesis straight A divided by straight E subscript 2 right parenthesis plus straight P left parenthesis straight E subscript 3 right parenthesis straight P left parenthesis straight A divided by straight E subscript 3 right parenthesis end fraction equals end cell row cell equals fraction numerator 2 over 5 left parenthesis 0.35 right parenthesis over denominator 2 over 5 left parenthesis 0.35 right parenthesis plus 1 fourth left parenthesis 0.20 right parenthesis plus 7 over 20 left parenthesis 0.10 right parenthesis end fraction equals fraction numerator 7 divided by 50 over denominator left parenthesis 7 divided by 50 right parenthesis plus left parenthesis 1 divided by 20 right parenthesis plus left parenthesis 7 divided by 200 right parenthesis end fraction end cell row cell equals fraction numerator 7 divided by 50 over denominator 9 divided by 40 end fraction equals 28 over 45 end cell end table end style

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28


Solution 29

Solution 30

Solution 31

Solution 32


Solution 33

Solution 34

Solution 35

Solution 36

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise MCQ

Solution 1

Correct option: (a)

  

 

Solution 2

Correct option: (a)

  

Solution 3

Correct option: (a)

  

 

Solution 4

Correct option: (b)

  

Solution 5

Correct option: (a)

  

Solution 6

Correct option: (a)

  

 

Solution 7

Correct option: (b)

  

Solution 8

Correct option: (d)

  

Solution 9

Correct option: (a)

  

Solution 10

Correct option: (c)

  

Solution 11

Correct option: (c)

  

Solution 12

Correct option: (d)

  

Solution 13

Correct option: (a)

  

 

Solution 14

Correct option: (a)

  

Solution 15

Correct option: (b)

  

 

Solution 16

Correct option: (a)

  

Solution 17

Correct option: (a)

  

Solution 18

Correct option: (c)

  

Solution 19

Correct option: (d)

  

Solution 20

Correct option: (a)

  

Solution 21

Correct option: (d)

  

Solution 22

Correct option: (d)

  

Solution 23

Correct option: (c)

  

Solution 24

Correct option: (b)

  

Solution 25

 

NOTE: Answer not matching with back answer. 

Solution 26

Correct option: (a)

  

 

Solution 27

Correct option: (c)

  

 

Solution 28

Correct option: (a)

  

Solution 29

Correct option: (b)

  

 

Solution 30

Correct option: (b)

Note: option is modified. 

 

Solution 31

Correct option: (a)

  

Solution 32

Correct option: (c)

  

 

Solution 33

Correct option: (c)

  

 

Solution 34

Correct option: (d)

  

Solution 35

Correct option: (d)

  

Solution 36

Correct option: (d)

  

 

Solution 37

Correct option: (d)

  

Solution 38

Correct option: (d)

  

Solution 39

Correct option: (a)

  

 

Solution 40

Correct option: (c)

  

Solution 41

Correct option: (c)

P(A B)= P(A) P(B) for independent events.

Solution 42

Correct option: (d)

  

Solution 43

Correct option: (c)

  

Solution 44

Correct option: (a)

  

 

Solution 45

Correct option: (b)

  

Solution 46

Correct option: (b)

  

Solution 47

Correct option: (c)

  

 

Solution 48

 

NOTE: Answer not matching with back answer. 

Solution 49

Correct option: (a)

  

 

Solution 50

Correct option: (a)

  

Solution 51

Correct option: (d)

  

Solution 52

Correct option: (d)

  

 

Solution 53

Correct option: (c)

 

 

Solution 54

Correct option: (b)

  

Solution 55

Correct option: (c)

  

Solution 56

Correct option: (c)

  

Solution 57

Correct option: (d)

  

Solution 58

Correct option: (d)

  

Solution 59

Correct option: (c)

 

NOTE: Answer not matching with back answer. 

RD Sharma Solution for Class 12 Science Mathematics Chapter 31 - Probability Page/Excercise 31VSAQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

TopperLearning provides step-by-step solutions for each question in each chapter in the RD Sharma textbook for class 9. Access the CBSE Class 9 Mathematics Chapter 31 - Probability for free. The textbook questions have been solved by our subject matter experts to help you understand how to answer them. Our RD Sharma Textbook Solutions will help you to study and revise, and you can easily clear your fundamentals of Chapter 31 - Probability.

Text Book Solutions

CBSE XII Science - Mathematics

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