Chapter 31 Probability Exercise Ex. 31.1

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Chapter 31 Probability Exercise Ex. 31.2

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6(i)

Solution 6(i)

Question 6(ii)

Solution 6(ii)

Question 6(iii)

Solution 6(iii)

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is probability that both drawn balls are black?Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Chapter 31 Probability Exercise Ex. 31.3

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5(i)

Solution 5(i)

Question 5(ii)

Solution 5(ii)

Question 5(iii)

Solution 5(iii)

Question 5(iv)

If A and B are two events such that

Solution 5(iv)

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

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

(ii) at least one is girl.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

Chapter 31 Probability Exercise Ex. 31.4

Question 1(i)

Solution 1(i)

Question 1(ii)

Solution 1(ii)

Question 1(iii)

Solution 1(iii)

Question 2

Solution 2

Question 3(i)

Solution 3(i)

Question 3(ii)

Solution 3(ii)

Question 3(iii)

Solution 3(iii)

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

begin mathsize 12px style table attributes columnalign left end attributes row cell text The   probabilities   of   two   students   end text straight A text   and   end text straight B text   coming   end text end cell row cell text to   the   school   in   time   are end text fraction numerator text 3 end text over denominator text 7 end text end fraction text   and   end text fraction numerator text 5 end text over denominator text 7 end text end fraction text   respectively .  end text end cell row cell text Assuming   that   the   events , ' end text straight A text   coming   in   time '  and  ' end text straight B end cell row cell text coming   in   time '  are   independent ,  find   the   probability   of   end text end cell row cell text only   one   of   them   coming   to   the   school   in   time .  end text end cell row cell text Write   atleast   one   advantage   of   coming   to   school   in   time. end text end cell end table end style

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.Question 24

Two dice are thrown together and the total score is noted. The event E, F and G are “a total 4”, “a total of 9 or more”, and “a total divisible by 5”, respectively. Calculate P (E), P(F) and P(G) and decide which pairs of events, if any, are independent.Solution 24

Question 25

Let A and B be two independent events such that P (A) = p1 and P (B) = p2. Describe in words the events whose probabilities are:

(i) p1p2 (ii) (1 – p1)p2 (iii) 1-(1- p1) (1 – p2) (iv) p1 + p2 = 2p1p2Solution 25

Chapter 31 Probability Exercise Ex. 31.5

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

Solution 14

Question 15

Solution 15

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

Solution 19

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

In a hockey match, both teams A and B scored same number of goals upto the end of the game, so to decide the winner, the refree asked both the captains to throw a die alternately and decide that the team, whose captain gets a first six, will be declared the winner. If the captain of team A was asked to start, find their respective probabilities of winning the match and state whether the decision of the refree was fair or not.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

Chapter 31 Probability Exercise Ex. 31.6

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

There machines E1, E2, E3 in a certain factory produce 50%, 25% and 25% respectively, of the total daily output of electric bulbs. It is known that 4% of the tubes produced one each of machines E1 and E2 are defective, and that 5% of those produced on E3 are defective. If one tube is picked up at random from a day’s production, calculate the probability that it is defective.Solution 13

Chapter 31 Probability Exercise Ex. 31.7

Question 1

Solution 1

Question 2

Solution 2

Question 3

Solution 3

Question 4

Solution 4

Question 5

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

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

Solution 9

Question 10

Solution 10

Question 11

Solution 11

Question 12

Solution 12

Question 13

Solution 13

Question 14

An item is manufactured by three machine A, B and C. out of the total number of items manufactured during a specified period, 50% are manufacture on machine A 30% on B and 20% on C. 2% of the items produced on A and 2% of items produced on B are defective and 3% of these produced on C are defective. All the items stored at one godown. One items is drawn at random and is found to be defective. What is the probability that it was manufactured on machine A?Solution 14

Question 15

There are three coins. One is two-headed coin (having head on both faces), another is biased coin that comes up heads 75% of the times and third is also a biased coin that comes up tail 40% of the times. One of the three coins is chosen at random and tossed, and it shows heads. What is the probability that it was the two-headed coin?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

Question 16

Solution 16

Question 17

Solution 17

Question 18

Solution 18

Question 19

In a group of 400 people, 160 are smokers and non-vegetarian, 100 are smokers and vegetarian and the remaining are non-smokers and vegetarian. The probabilities of getting a special chest disease are 35%, 20% and 10% respectively. A person is chosen from the group at random and is found to be suffering from the disease. What is the probability that the selected person is a smoker and non-vegetarian?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

Question 20

Solution 20

Question 21

Solution 21

Question 22

Solution 22

Question 23

Solution 23

Question 24

Solution 24

Question 25

Solution 25

Question 26

Solution 26

Question 27

Solution 27

Question 28

Solution 28

Question 29

Solution 29

Question 30

Solution 30

Question 31

Solution 31

Question 32

Solution 32

Question 33

Solution 33

Question 34

Solution 34

Question 35

Solution 35

Question 36

Solution 36


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