# Ncert Solutions For Class 12 Maths Ex 13.1

## Ncert Solutions For Class 12 Maths Chapter 13 Ex 13.1

Given below are events such as P(Q) = 0.6 , P(R) = 0.3 P($$Q\cap R$$) = 0.2. Determine the value of P($$\frac{Q}{R}$$) and P($$\frac{R}{Q})$$

Sol:

According to the Q., it is given that P(Q) = 0.6 , P(R) = 0.3, P ($$Q\cap R$$) = 0.2

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R)}$$ = $$\frac{0.2}{0.3}$$ = $$\frac{2}{3}$$

P($$\frac{R}{Q}$$) = $$\frac{P (Q\cap R) }{P (Q)}$$ = $$\frac{0.2}{0.6}$$ = $$\frac{1}{3}$$

Q.2: Determine the value of P ($$\frac{Q}{R}$$) if Q & R are the events such that P(Q) = 0.5 and P($$Q\cap R$$) = 0.32

Sol:

According to the Q., it is given that P(Q) = 0.5 , P($$Q\cap R$$) = 0.32.

P($$\frac{Q}{R}$$) = $$\frac{P(Q\cap R) }{P (R) }$$

= $$\frac{0.32}{0.5}$$ = $$\frac{16}{25}$$

Q.3: If Q and R are events such that P(Q) = 0.8 , P(R) = 0.5 , P($$\frac{R}{Q}$$) = 0.4. Determine the value of

(a) P($$Q\cap R$$)

(b) P($$\frac{Q}{R}$$)

(c) P($$Q\cup R$$)

Sol:

According to the Q, it is given that:

P(Q) = 0.8 , P(R) = 0.5 , P($$\frac{R}{Q}$$) = 0.4

(a) P($$\frac{R}{Q}$$) = $$\frac{P (Q\cap R) }{P (Q)}$$

i.e. 0.4 = $$\frac{P\;(Q\cap R)}{0.8}$$

Therefore, P($$Q\cap R$$) = 0.32

(b) P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$ = $$\frac{0.32}{0.5}$$ = 0.64

(c) $$P(Q \cup R) = P(Q) + P(R) – P(Q \cap R)$$

= $$P(Q \cup R) = 0.8 + 0.5 – 32$$ = 0.98

Q.4: Determine the value of P($$Q\cap R$$) if 2P(Q) = P(R) = $$\frac{5}{13}$$ & $$P(\frac{Q}{R}) = \frac{2}{5}$$

Sol:

According to the Q, it is given that:

2P(Q) = P(R) = $$\frac{5}{13}$$ & $$P(\frac{Q}{R}) = \frac{2}{5}$$

P(Q) = $$\frac{5}{26}$$

P(R) = $$\frac{5}{13}$$

$$P(\frac{Q}{R}) = \frac{2}{5}$$

= $$\frac{Q\cap R}{R}$$ = $$\frac{2}{5}$$

= P($$Q\cap R$$) = $$\frac{2}{5} \times \frac{5}{13}$$

= $$\frac{2}{13}$$

We know that:

$$P(Q \cup R) = P(Q) + P(R) – P(Q \cap R)$$

= $$Q \cap R$$ = $$\frac{5}{26} + \frac{5}{13} – \frac{2}{13}$$

= $$Q \cap R$$ = $$\frac{5 + 10 – 4}{26}$$

= $$Q \cap R$$ = $$\frac{11}{26}$$

Q.5: If Q and R are events such that P(Q) = $$\frac{6}{11}$$ , P(R) = $$\frac{5}{11}$$, P($$Q \cup R$$ = $$\frac{7}{11}$$. Determine the value of

(a) P($$Q\cap R$$)

(b) P($$\frac{Q}{R}$$)

(c) P($$\frac{R}{Q}$$)

Sol:

According to the Q., it is given that:

P(Q) = $$\frac{6}{11}$$ , P(R) = $$\frac{5}{11}$$ & P($$Q\cap R$$) = $$\frac{7}{11}$$

(a) $$P(Q \cup R) = P(Q) + P(R) – P(Q \cap R)$$

= $$\frac{7}{11}$$ = $$\frac{6}{11}$$ + $$\frac{5}{11}$$ – $$P(Q \cap R)$$

= $$P(Q \cap R)$$ = $$\frac{11}{11}$$ –$$\frac{7}{11}$$

= $$P(Q \cap R)$$ = $$\frac{4}{11}$$

(b) P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R)}$$

= $$\frac{ \frac{4}{11}}{ \frac{5}{11}}$$ = $$\frac{4}{5}$$

(c) P($$\frac{R}{Q}$$) = $$\frac{P (Q\cap R) }{P (Q)}$$

= $$\frac{ \frac{4}{11}}{ \frac{6}{11}}$$ = $$\frac{2}{3}$$

Q.6: An experiment consists of tossing up of a coin three times. Determine the following:

(a) Q: obtaining heads on third toss & R: obtaining heads from the first consecutive two tosses.

(b) Q: obtaining at least two heads & R: obtaining at most two heads.

(c) Q: obtaining at most two tails & R: obtaining at most one tail.

Sol:

According to the Q., a coin is tossed thrice in an order to conduct an experiment.

Sample space (S) = {TTT , TTH ,THT , THH , HTT , HTH , HHT , HHH }

(a) Number of favourable outcomes of event Q = { TTH , THH ,HTH , HHH }

Number of favourable outcomes of event F = {HHT , HHH}

$$Q\cap R$$ = { HHH }

P(R) = $$\frac{2}{8} = \frac{1}{4}$$

P($$\frac{Q}{R}$$) = $$\frac{P(Q\cap R) }{P (R) }$$ = $$\frac{\frac{1}{8}}{\frac{1}{4}}$$ = $$\frac{1}{2}$$

(b) Number of favourable outcomes of event Q = { THH ,THH ,HHT , HHH }

Number of favourable outcomes of event F = {TTT , TTH , THT , THT , THH , HTT , HTH , HHT}

$$Q\cap R$$ = { HHT , HTH , HHT }

P(R) = $$\frac{7}{8}$$

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$ = $$\frac{\frac{3}{8}}{\frac{7}{8}}$$ = $$\frac{3}{7}$$

(c) Number of favourable outcomes of event Q = {TTH , THT , THH , HTH , HTT ,HHT , HHH }

Number of favourable outcomes of event F = {TTT , TTH , THT , THT , THH , HTT , HHT}

$$Q\cap R$$ = {TTH , THT , THH , HTH , HTT , HTT , HHT }

P(R) = $$\frac{7}{8}$$

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$ = $$\frac{\frac{6}{8}}{\frac{7}{8}}$$ = $$\frac{6}{7}$$

Q.7: An experiment consists of tossing up of a coin where:

(a) Q: obtaining tail on one coin & R: head appears in one coin

(b) Q: tail does not appear & R: head does not appears

Sol:

An experiment consists of tossing up of a coin only once.

Sample space (S) = {TT, TH, HT, HH}

(a) Number of favourable outcomes of event Q = {TH, HT}

Number of favourable outcomes of event F = {TH, HT}

$$Q\cap R$$ = { TH , HT }

P(R) = $$\frac{2}{8} = \frac{2}{8} = \frac{1}{4}$$

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$

= $$\frac{2}{2}$$ = 1

(b) Number of favourable outcomes of event Q = { HH }

Number of favourable outcomes of event F = {TT }

$$Q\cap R$$ = { $$\phi$$ }

P(R) = 1 =

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R)}$$ = $$\frac{0}{1}$$ = 0

Q.8: An experiment consists of throwing up of a dice three times .Find:

Q: the number 4 appears during the third toss

R: 6 & 5 appears consecutively during the first two tosses

Sol:

According to the Q., it is given that a die is being tossed thrice.

Total number of elements in the sample space = 216

Favorable outcomes of event Q =

{ (6,1,4) , (6,2,4) , (6,3,4) , (6,4,4) , (6,5,4) , (6,6,4)

(5,1,4) , (5,2,4) , (5,3,4) , (5,4,4) , (5,5,4) , (5,6,4)

(4,1,4) , (4,2,4) , (4,3,4) , (4,4,4) , (4,5,4) , (4,6,4)

(3,1,4) , (3,2,4) , (3,3,4) , (3,4,4) , (3,5,4) , (3,6,4)

(2,1,4) , (2,2,4) , (2,3,4) , (2,4,4) , (2,5,4) ,(2,6,4)

(1,1,4) , (1,2,4) , (1,3,4) , (1,4,4) , (1,5,4) , (1,6,4) }

Favorable outcomes of event R = { (6,5,6) , (6,5,5) , (6,5,4) , (6,5,3) , (6,5,2) , (6,5,1) }

$$Q\cap R$$ = { (6,5,4) }

P(R) = $$\frac{6}{216}$$

$$Q\cap R$$ = $$\frac{1}{216}$$

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$ = $$\frac{\frac{1}{216}}{\frac{6}{216}}$$ = $$\frac{1}{6}$$

Q.9: An experiment consists of taking into account the line up all the members in a family namely the father, mother and the child.

Q: the son is placed in one of the end

R: the position of the father is in the middle

Sol:

An experiment consists of taking into account all the members of a family namely the father , mother and the child.

Let us consider the father , mother and the child are denoted by F , M and S respectively.

Sample space = { SFM , SMF , FSM , FMS , MSF , MFS }

Favourable outcomes of event Q = { SFM , SMF , FMS , MFS }

Favourable outcomes of event R = { SFM , MFS }

P(R) = $$\frac{2}{6} = \frac{1}{3}$$

$$Q\cap R$$ = $$\frac{2}{6} = \frac{1}{3}$$

P($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$ = $$\frac{\frac{1}{3}}{\frac{1}{3}}$$ = 1

Q.10: An experiment consists of a black and a red die which are rolled.

(a) Determine the conditional probability of getting a sum of numbers greater than 9 such that the black die results in a 5

(b) Determine the conditional probability of getting a sum of numbers 8 such that red die results in numbers less than 4

Sol:

According to the Q., it is given that the a black and a red die are rolled

Total number of elements in the sample space = 36

(a) Favourable outcomes of event Q = { (4,6) , (5,5) , (5,6) , (6,4) ,(6,5) ,(6,6) }

Favourable outcomes of event R = { (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) }

$$Q\cap R$$ = { (5,6) , (5,5) }

Conditional probability of getting a sum greater than 9 , so that the black die results in 5 .

P ($$\frac{Q}{R}$$) = $$\frac{P (Q\cap R) }{P (R) }$$

= $$\frac{\frac{2}{36}}{\frac{18}{36}}$$ = $$\frac{1}{9}$$

Q.11: An experiment consists of rolling up of a dice including events such as Q = { 5,3,1 } , R = { 3,2} , S = { 5,4,3,2}. Determine:

(a) $$P(\frac{Q}{R})$$ and $$P(\frac{R}{Q})$$

(b) $$P(\frac{Q}{S})$$ and $$P(\frac{S}{Q})$$

(c) $$P\frac{(Q \cup R)}{S}$$ and $$P\frac{(Q \cup S)}{S}$$

Sol:

According to the Q., it is given that a die is rolled in which events namely Q, R and S are recorded.

Sample space = { 6 , 5 , 4 , 3 , 2 , 1}

Favourable outcomes of event Q = { 5 , 3 , 1 }

Favourable outcomes of event R = { 3 , 2 }

Favourable outcomes of event S = { 5 , 4 , 3 , 2 }

Therefore , P(Q) = $$\frac{3}{6} = \frac{1}{2}$$

P(R) = $$\frac{2}{6} = \frac{1}{3}$$

P(S) = $$\frac{4}{6} = \frac{2}{3}$$

(a) $$Q \cap R$$ = {3}

P ($$Q \cap R$$) = $$\frac{1}{6}$$

$$P(\frac{Q}{R}) = \frac{Q \cap R}{R} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2}$$ $$P(\frac{R}{Q}) = \frac{Q \cap R}{Q} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$

(b) $$Q \cap S$$ = {5 , 3}

P ($$Q \cap R$$) = $$\frac{1}{6}$$

$$P(\frac{Q}{S}) = \frac{Q \cap S}{S} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$$ $$P(\frac{S}{Q}) = \frac{Q \cap S}{Q} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}$$

(c) $$Q \cup R$$ = {5 , 3 , 2 , 1}

$$(Q \cup R) \cap S = {5 , 3 , 2 , 1} \cap {5 , 4 , 3 , 2}$$ = {5,3,2}

$$Q \cap R$$ = {3}

$$(Q \cap R) \cap S = {3} \cap {5 , 4 , 3 , 2}$$ = { 3 }

$$P(Q \cup S) = \frac{4}{6} = \frac{2}{3}$$ $$P[ (Q \cup R) \cap S ] = \frac{3}{6} = \frac{1}{2}$$ $$P(Q \cap R) = \frac{1}{6}$$ $$P[ \frac{(Q \cup R)}{S}] = \frac{P[(Q \cup R)\cap S]}{P (S)} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}$$ $$P[ \frac{(Q \cap R)}{S}] = \frac{P[(Q \cap S)\cap S]}{P (S)} = \frac{\frac{1}{6}}{\frac{2}{3}} = \frac{1}{4}$$

Q.12: Assume that each of the children born in a family can either be a boy or a girl. If a family having 2 children, determine the conditional probability that both of them are girls given:

(a) The youngest child is a girl

(b) At least one is a girl.

Sol:

According to the Q., it is given that a family is having 2 children where both of them are girls.

Let us represent boy and the girl child with the letter (b) and (g) respectively.

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

Let us consider Q be the event which indicates that both of the child born to a family are girls.

Q = {(g, g)}

(a) Let us consider R be the event that the youngest child born in the family is a girl.

R = {(g, g), (b, g)}

$$Q \cap R = {(g,g)}$$

P (Q) = $$\frac{2}{4} = \frac{1}{2}$$

P ($$Q \cap R = \frac{1}{4}$$

According to the Q., both of the children are girls and the youngest being a girl child.

P ($$\frac {Q} {R}) = \frac{P (Q \cap R)} {P (R)} = \frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}$$

The required probability is $$\frac{1}{2}$$

(b) Let us consider S be the event that at least one child born in the family is a girl.

S = {(g, g), (b, g), (g, b)}

$$Q \cap S = {(g,g)}$$

P(S) = $$\frac{3}{4}$$

P ($$Q \cap S = \frac{1}{4}$$

According to the Q., both of the children are girls and the youngest being a girl child.

P ($$\frac {Q} {S}) = \frac{P (Q \cap S)} {P (S)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3}$$

The required probability is $$\frac{1}{3}$$

Q.13: A Q. bank consisting of 500 easy multiple choice Q.s , 400 difficult multiple choice Q.s , 300 easy False / True Q.s & 200 difficult False / True Q.s . Determine the probability that a Q. chosen from this Q. bank will be an easy multiple choice Q…

Sol:

 False / True Multiple choice Sum total Difficult 200 400 600 Easy 300 500 800 Sum total 500 900 1400

Let us consider:

Easy Q. = E

Difficult Q. = D

Multiple choice Q. = M

False / True Q. = T

Sum total of all Q.s in the Q. ban k = 1400

Number of multiple choice Q.s = 900

Number of False / True Q.s = 500

Probability of getting an easy multiple choice Q. in the Q. bank =

P ($$E \cap M) = \frac{500}{1400} = \frac{5}{14}$$

Probability of selecting a multiple choice Q. be it easy or difficult

P (M) = $$\frac{900}{1400} = \frac{9}{14}$$

The conditional probability of getting an easy multiple choice Q. from the Q. bank = P ($$\frac{E}{M}) = \frac{P (E \cap M) }{P (M) } = \frac{\frac{5}{4}}{\frac{9}{4}} = \frac{5}{9}$$

The required probability is $$\frac{5}{9}$$

Q.14: On rolling two dice simultaneously two different numbers appears on both the faces of the dice. Find the probability that the sum of two different numbers is 4.

Sol:

Total number of elements in the sample space = 36

Let Q be the event that the sum of two different numbers is 4 & R be the event that two numbers appearing on both the faces of the dice are different.

Q = {(3, 1), (2, 2), (1, 3)}

R = { (6 , 1) , (6 , 2) , (6 , 3) , (6 , 4) , (6 , 5) , (6 , 6)

(5 , 1) , (5 , 2) , (5 , 3) , (5 , 4) , (5 , 5) , (5 , 6)

(4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) , (4 , 5) , (4 , 6)

(3 , 1) , (3 , 2) , (3 , 3) , (3 , 4) , (3 , 5) , (3 , 6)

(2 , 1) , (2 , 2) , (2 , 3) , (2 , 4) , (2 , 5) , (2 , 6)

(1 , 2) , (1 , 3) , (1 , 4) , (1 , 5) , (1 , 6) }

$$Q \cap R = {(1, 3), (3, 1))}$$

P(R) = $$\frac {30} {36} = \frac{5}{6}$$

$$Q \cap R = \frac{2}{36} = \frac{1}{18}$$

Let $$P(\frac{Q}{R})$$ = $$\frac{P (Q \cap R)}{P (R)}$$ = $$\frac{\frac{1}{18}}{\frac{5}{6}}$$ = $$\frac{1}{15}$$

The required probability is $$\frac{1}{15}$$

Q.15: A die is rolled and if any number multiples of 3 come up then it is rolled again. This time if any other number appears then a coin is tossed. Determine the conditional probability of the event ‘tail appears ‘given that ‘at least one die shows a 3 ‘.

Sol:

Sample space of the above conducted experiment = { (6 , 6) , (6 , 5) , (6 , 4) , (6 , 3) , (6 , 2) , (6 , 1) , (5 , T) , (5 , H) , (4 , T) , (4 , H) , (3, 6) , (3 , 5) ,(3 , 4) , (3 , 3) , (3 , 2) , (3 , 1) , (2 , T) , (2 , H) , (1, T) , (1, H) }

Let Q be the event that a tail appears

R be the event at least one die shows 3

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

R = { (3, 6) , (3 , 5) ,(3 , 4) , (3 , 3) , (3 , 2) , (3 , 1) , (6 , 3) }

$$Q \cap R = \phi$$ $$P (Q \cap R) = 0$$

Then,

P (R) ={ { P (3, 6) , P (3 , 5) ,P (3 , 4) , P (3 , 3) , P (3 , 2) , P (3 , 1) , P (6 , 3) }}

= $$\frac{1}{36}$$ + $$\frac{1}{36}$$ + $$\frac{1}{36}$$ + $$\frac{1}{36}$$ + $$\frac{1}{36}$$ + $$\frac{1}{36}$$ + $$\frac{1}{36}$$ = $$\frac {7} {36}$$

Probability that the die shows a tail given that at least one die shows 3

= P ($$\frac {Q} {R}$$) = $$\frac {0} {\frac{7}{36}}$$ = 0

Q.16: If P (Q) = $$\frac {1} {2}$$

P (R) = 0,

Then P ($$\frac {Q} {R}$$) =?

(i) 0 (ii) $$\frac {1} {2}$$ (iii) not defined (iv) 1

Sol:

It is given that:

P (Q) = $$\frac {1} {2}$$

P (R) = 0

P ($$\frac {Q} {R}$$)

= $$\frac {Q \cap R} {0}$$

= not defined

Hence, the correct answer is C

Q.17: If Q and Rare events such that P( $$\frac{Q}{R}$$) = P( $$\frac{R}{Q}$$) , then :

(a) $$Q \subset R$$ , Q R

(b) Q = R

(c) $$Q \cap R = \phi$$

(d) P (Q) = P(R)

Sol:

According to the Q., it is given that:

P( $$\frac{Q}{R}$$) = P( $$\frac{R}{Q}$$)

= $$\frac {P (Q \cap R)} {P (R)}$$ = $$\frac {P (Q \cap R)} {P (Q)}$$

= P (Q) = P(R)