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)

The correct answer is D

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