NCERT Solutions For Class 12 Maths Chapter 13 Exercise 13.2

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Chapter 13 Ex.13.2 Question 1

If \({\text{P}}\left( {\text{A}} \right) = \frac{3}{5}{\text{ and P}}\left( {\text{B}} \right) = \frac{1}{5}\), find \({\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\)if A and B are independent events.

Solution

Given, \({\text{P}}\left( {\text{A}} \right) = \frac{3}{5}\) and \({\text{P}}\left( {\text{B}} \right) = \frac{1}{5}\)

A and B are independent events.

Thus, \({\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{3}{5} \times \frac{1}{5} = \frac{3}{{25}}\)

Chapter 13 Ex.13.2 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.

Solution

Since there are \(26\) black cards in a deck of \(52\) cards.

Let \({\text{P}}\left( {\text{A}} \right)\)be the probability of getting a black card in the first draw.

\(\therefore {\text{ P}}\left( {\text{A}} \right) = \frac{{26}}{{52}}\; = \frac{1}{2}\)

Let \({\text{P}}\left( {\text{B}} \right)\)be the probability of getting a black card on second draw.

Since the card is not replaced,

\(\therefore {\text{ P}}\left( {\text{B}} \right) = \frac{{25}}{{51}}\)

Therefore, probability of getting both the cards black\( = \frac{1}{2} \times \frac{{25}}{{51}} = \frac{{25}}{{102}}\)

Chapter 13 Ex.13.2 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.

Solution

Let A, B, and C be the respective events that the first, second, and the third drawn orange is good.

Thus, probability that first drawn orange is good, \({\text{P}}\left( {\text{A}} \right) = \frac{{12}}{{15}}\)

The oranges are not replaced.

Thus, probability of getting second orange good, \({\text{P}}\left( {\text{B}} \right) = \frac{{11}}{{14}}\)

Similarly, probability of getting third orange good, \({\text{P}}\left( {\text{C}} \right) = \frac{{10}}{{13}}\)

Since the box is approved for sale, if all the three oranges are good.

Thus, probability of getting all the oranges good \( = \frac{{12}}{{15}} \times \frac{{11}}{{14}} \times \frac{{10}}{{13}} = \frac{{44}}{{91}}\)

Thus, the probability that the box is approved for sale is \( = \frac{{44}}{{91}}\)

Chapter 13 Ex.13.2 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.

Solution

The sample space is given by, \({\text{S = }}\left\{ \begin{align}{l}\left( {{\text{H,1}}} \right),\left( {{\text{H,2}}} \right),\left( {{\text{H,3}}} \right),\left( {{\text{H,4}}} \right),\left( {{\text{H,5}}} \right),\left( {{\text{H,6}}} \right)\\\left( {{\text{T,1}}} \right),\left( {{\text{T,2}}} \right),\left( {{\text{T,3}}} \right),\left( {{\text{T,4}}} \right),\left( {{\text{T,5}}} \right),\left( {{\text{T,6}}} \right)\end{align} \right\}\)

Let A: Head appears on the coin

\[\begin{align}&{\text{A = }}\left\{ {\left( {{\text{H,1}}} \right),\left( {{\text{H,2}}} \right),\left( {{\text{H,3}}} \right),\left( {{\text{H,4}}} \right),\left( {{\text{H,5}}} \right),\left( {{\text{H,6}}} \right)} \right\}\\ &\Rightarrow {\text{P}}\left( {\text{A}} \right) = \frac{6}{{12}} = \frac{1}{2}\end{align}\]

B: \(3\) on die\( = \left\{ {\left( {{\text{H,3}}} \right),\left( {{\text{T,3}}} \right)} \right\}\)

\[\begin{align}& \Rightarrow\; {\text{P}}\left( {\text{B}} \right) = \frac{2}{{12}} = \frac{1}{6}\\&{\text{A}} \cap {\text{B}} = \left\{ {\left( {{\text{H,3}}} \right)} \right\}\\&{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{{12}}\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{2}{6} = {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\end{align}\]

Therefore, A and B are independent events.

Chapter 13 Ex.13.2 Question 5

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

Solution

The sample space (S) is \({\text{S}} = \left\{ {1,2,3,4,5,6} \right\}\)

\[\begin{align}&{\text{Let A: the number is even}} = \left\{ {2,4,6} \right\}\\ &\Rightarrow\; \; {\text{P}}\left( {\text{A}} \right) = \frac{3}{6} = \frac{1}{2}\\&{\text{B: the number is red}} = \left\{ {1,2,3} \right\}\\& \Rightarrow\; \; {\text{P}}\left( {\text{B}} \right) = \frac{3}{6} = \frac{1}{2}\\&\therefore {\text{A}} \cap {\text{B}} = \left\{ {\text{2}} \right\}\end{align}\]

\[\begin{align}&{\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{6}\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \ne \frac{1}{6}\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) \ne {\text{P}}\left( {{\text{AB}}} \right)\end{align}\]

Therefore, A and B are not independent events.

Chapter 13 Ex.13.2 Question 6

Let E and F be the events with \({\text{P}}\left( {\text{E}} \right) = \frac{3}{5}{\text{, P}}\left( {\text{F}} \right) = \frac{3}{{10}}{\text{ and P}}\left( {{\text{E}} \cap {\text{F}}} \right) = \frac{1}{5}\). Are E and F are independent?

Solution

Given \({\text{P}}\left( {\text{E}} \right) = \frac{3}{5}{\text{, P}}\left( {\text{F}} \right) = \frac{3}{{10}}\) and \({\text{P}}\left( {{\text{EF}}} \right) = {\text{P}}\left( {{\text{E}} \cap {\text{F}}} \right) = \frac{1}{5}\)

\[\begin{align}&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = \frac{3}{5} \times \frac{3}{{10}} = \frac{9}{{50}} \ne \frac{1}{5}\\&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) \ne {\text{P}}\left( {{\text{E}} \cap {\text{F}}} \right)\end{align}\]

Therefore, E and F are not independent.

Chapter 13 Ex.13.2 Question 7

Given that the events A and B are such that \({\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{3}{5}\) and \({\text{P}}\left( {\text{B}} \right) = {\text{p}}\). Find p if they are (i) mutually exclusive (ii) independent.

Solution

Given, \({\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{3}{5}\) and \({\text{P}}\left( {\text{B}} \right) = {\text{p}}\)

(i) When A and B are mutually exclusive, \({\text{A}} \cap {\text{B}} = \phi \)

\[\begin{align}&\therefore {\text{ P}}\left( {{\text{A}} \cap {\text{B}}} \right) = 0\\&{\text{Since, P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &\Rightarrow\; \; \frac{3}{5} = \frac{1}{2} + {\text{p}} - 0\\ &\Rightarrow\; \; {\text{p}} = \frac{3}{5} - \frac{1}{2} = \frac{1}{{10}}\end{align}\]

(ii) When A and B are independent,\({\text{ P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2}{\text{p}}\)

\[\begin{align}{\text{Since, P}}\left( {{\text{A}} \cup {\text{B}}} \right)& = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &\Rightarrow\; \; \frac{3}{5} = \frac{1}{2} + {\text{p}} - \frac{1}{2}{\text{p}}\\ &\Rightarrow\; \; \frac{3}{5} = \frac{1}{2} + \frac{{\text{p}}}{2}\\ &\Rightarrow\; \; \frac{{\text{p}}}{2} = \frac{3}{5} - \frac{1}{2} = \frac{1}{{10}}\\ &\Rightarrow\; \; \frac{{\text{p}}}{2} = \frac{2}{{10}} = \frac{1}{5}\end{align}\]

Chapter 13 Ex.13.2 Question 8

Let \(A\) and \(B\) be independent events with \({\text{P}}\left( {\text{A}} \right) = 0.3\) and \({\text{P}}\left( {\text{B}} \right) = 0.4\). Find

\(\begin{align}&\left( {\text{i}} \right){\text{ P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\&\left( {{\text{ii}}} \right){\text{ P}}\left( {{\text{A}} \cup {\text{B}}} \right)\\&\left( {{\text{iii}}} \right){\text{ P}}\left( {\left. {\text{A}} \right|{\text{B}}} \right)\\&\left( {{\text{iv}}} \right){\text{ P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right)\end{align}\)

Solution

Given, \({\text{P}}\left( {\text{A}} \right) = 0.3\) and \({\text{P}}\left( {\text{B}} \right) = 0.4\).

If \(A\) and \(B\) are independent events, then

\[\begin{align}{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) &= {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &= 0.3 \times 0.4\\& = 0.12\end{align}\]

Since, \({\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\)

\( \Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = 0.3 + 0.4 - 0.12 = 0.58\)

Since, \({\text{P}}\left( {\left. {\text{A}} \right|{\text{B}}} \right) = \frac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}}\)

\( \Rightarrow\; \; {\text{P}}\left( {\left. {\text{A}} \right|{\text{B}}} \right) = \frac{{0.12}}{{0.4}} = 0.3\)

Since, \({\text{P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right) = \frac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}\)

\( \Rightarrow\; \; {\text{P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right) = \frac{{0.12}}{{0.3}} = 0.4\)

Chapter 13 Ex.13.2 Question 9

If A and B are two events such that \({\text{P}}\left( {\text{A}} \right) = \frac{1}{4}{\text{, P}}\left( {\text{B}} \right) = \frac{1}{2}\)and \({\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{8}\), find P (not A and not B).

Solution

Given, \({\text{P}}\left( {\text{A}} \right) = \frac{1}{4}{\text{, P}}\left( {\text{B}} \right) = \frac{1}{2}\) and \({\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{8}\).

\[\begin{align}&{\text{P (not on A and not on B)}} = {\text{P}}\left( {{\text{A'}} \cap {\text{B'}}} \right)\\&{\text{P (not on A and not on B)}} = {\text{P}}\left( {{{\left( {{\text{A}} \cup {\text{B}}} \right)}^\prime }} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{A'}} \cap {\text{B' = }}\left( {{\text{A}} \cup {\text{B}}} \right){\text{'}}} \right]\\ &= 1 - {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)\\ &= 1 - \left[ {{\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)} \right]\\& = 1 - \left[ {\frac{1}{4} + \frac{1}{2} - \frac{1}{8}} \right]\\ &= 1 - \frac{5}{8}\\ &= \frac{3}{8}\end{align}\]

Chapter 13 Ex.13.2 Question 10

Events A and B are such that \({\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {\text{B}} \right) = \frac{7}{{12}}\) and P (not A or not B) \( = \frac{1}{4}\). State whether A and B are independent?

Solution

Given, \({\text{P}}\left( {\text{A}} \right) = \frac{1}{2}{\text{, P}}\left( {\text{B}} \right) = \frac{7}{{12}}\)and P (not A or not B)\( = \frac{1}{4}\)

\[\begin{align} &\Rightarrow\; \; {\text{P}}\left( {{\text{A'}} \cup {\text{B'}}} \right) = \frac{1}{4}\\& \Rightarrow\; \; {\text{P}}\left( {{{\left( {{\text{A}} \cap {\text{B}}} \right)}^\prime }} \right) = \frac{1}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{A'}} \cup {\text{B' = }}\left( {{\text{A}} \cap {\text{B}}} \right){\text{'}}} \right]\end{align}\]

\[\begin{align}& \Rightarrow\; \; 1 - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{1}{4}\\ &\Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = \frac{3}{4}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;.....\;\left( 1 \right)\\&{\text{But}},\;{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{7}{{12}} = \frac{7}{{24}}\;\;\;\;\;.....\;\left( 2 \right)\\&{\text{Here,}}\;\frac{3}{4} \ne \frac{7}{{24}}\\&\therefore \;{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) \ne {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\end{align}\]

Therefore, A and B are not independent events.

Chapter 13 Ex.13.2 Question 11

Given two independent events \(A\) and \(B\) such \({\text{P}}\left( {\text{A}} \right) = 0.3,\;{\text{P}}\left( {\text{B}} \right) = 0.6\). Find

(i) \({\text{P}}\left( {{\text{A and B}}} \right)\)

(ii) \({\text{P}}\left( {{\text{A and not B}}} \right)\)

(iii) \({\text{P}}\left( {{\text{A or B}}} \right)\)

(iv) \({\text{P}}\left( {{\text{neither}}\;{\text{A nor B}}} \right)\)

Solution

Given, \({\text{P}}\left( {\text{A}} \right) = 0.3\;{\text{and}}\;{\text{P}}\left( {\text{B}} \right) = 0.6\)

\(A\) and \(B\) are independent events.

(i)\({\text{P}}\left( {{\text{A and B}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\)

\( \Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = 0.3 \times 0.6 = 0.18\)

(ii)\({\text{P}}\left( {{\text{A and not B}}} \right) = {\text{P}}\left( {{\text{A}} \cap {\text{B'}}} \right)\)

\[\begin{align} &= {\text{P}}\left( {\text{A}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &= 0.3 - 0.18\\ &= 0.12\end{align}\]

(iii)\({\text{P}}\left( {{\text{A or B}}} \right) = {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)\)

\[\begin{align}& = {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)\\ &= 0.3 + 0.6 - 0.18\\ &= 0.72\end{align}\]

(iv)\({\text{P}}\left( {{\text{neither}}\;{\text{A nor B}}} \right) = {\text{P}}\left( {{\text{A'}} \cap {\text{B'}}} \right)\)

\[\begin{align} &= {\text{P}}\left( {{{\left( {{\text{A}} \cup {\text{B}}} \right)}^\prime }} \right) \qquad \left[ {{\text{A'}} \cap {\text{B'}} = \left( {{\text{A}} \cup {\text{B}}} \right){\text{'}}} \right]\\ &= 1 - {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)\\& = 1 - 0.72\\ &= 0.28\end{align}\]

Chapter 13 Ex.13.2 Question 12

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

Solution

Probability of getting an odd number in a single throw of a die \( = \frac{3}{6} = \frac{1}{2}\)

Similarly, probability of getting an even number\( = \frac{3}{6} = \frac{1}{2}\)

Probability of getting an even number three times \( = \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\)

\(=1 −\) Probability of getting an odd number in none of the throws

\(=1 −\) Probability of getting an even number thrice

\[\begin{align} = 1 - \frac{1}{8}\\ = \frac{7}{8}\end{align}\]

Chapter 13 Ex.13.2 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.

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

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

Solution

Given,

Total number of balls\( = 18\)

Number of red balls\(= 8\)

Number of black balls\( = 10\)

(i) Probability of getting a red ball in the first draw \( = \frac{8}{{18}} = \frac{4}{9}\)

The ball is replaced after the first draw.

\(∴\) Probability of getting a red ball in the second draw \( = \frac{8}{{18}} = \frac{4}{9}\)

Therefore, probability of getting both the balls red \( = \frac{4}{9} \times \frac{4}{9} = \frac{{16}}{{81}}\)

(ii) Probability of getting first ball black\( = \frac{{10}}{{18}} = \frac{5}{9}\)

The ball is replaced after the first draw.

Probability of getting second ball as red  \( = \frac{8}{{18}} = \frac{4}{9}\)

Therefore, probability of getting first ball as black and second ball as red =\( = \frac{5}{9} \times \frac{4}{9} = \frac{{20}}{{81}}\)

(iii) Probability of getting first ball as red\( = \frac{8}{{18}} = \frac{4}{9}\)

The ball is replaced after the first draw.

Probability of getting second ball as black \( = \frac{{10}}{{18}} = \frac{5}{9}\)

Therefore, probability of getting first ball as black and second ball as red \( = \frac{4}{9} \times \frac{5}{9} = \frac{{20}}{{81}}\)

Therefore, probability that one of them is black and other is red \(=\) Probability of getting first ball black and second as red \(+\) Probability of getting first ball red and second ball black \( = \frac{{20}}{{81}} + \frac{{20}}{{81}} = \frac{{40}}{{81}}\)

Chapter 13 Ex.13.2 Question 14

Probability of solving specific problem independently by A and B are \(\frac{1}{2}\) and \(\)respectively.If both try to solve the problem independently, find the probability that

(i) the problem is solved

(ii) exactly one of them solves the problem.

Solution

Probability of solving the problem by A, \({\text{P}}\left( {\text{A}} \right) = \frac{1}{2}\)

Probability of solving the problem by B, \({\text{P}}\left( {\text{B}} \right) = \frac{1}{3}\)

Since the problem is solved independently by A and B,

\[\begin{align}&{\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}\\&{\text{P}}\left( {{\text{A'}}} \right) = 1 - {\text{P}}\left( {\text{A}} \right) = 1 - \frac{1}{2} = \frac{1}{2}\\&{\text{P}}\left( {{\text{B'}}} \right) = 1 - {\text{P}}\left( {\text{B}} \right) = 1 - \frac{1}{3} = \frac{2}{3}\end{align}\]

(i)Probability that the problem is solved\( = {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right)\)

\[\begin{align} &= {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{AB}}} \right)\\ &= \frac{1}{2} + \frac{1}{3} - \frac{1}{6}\\& = \frac{4}{6} = \frac{2}{3}\end{align}\]

(ii) Probability that exactly one of them solves the problem \({\text{ = P}}\left( {\text{A}} \right) \cdot {\text{P}}\left( {{\text{B'}}} \right) + {\text{P}}\left( {\text{B}} \right) \cdot {\text{P}}\left( {{\text{A'}}} \right)\)

\[\begin{align} &= \frac{1}{2} \times \frac{2}{3} + \frac{1}{2} \times \frac{1}{3}\\ &= \frac{1}{3} + \frac{1}{6} = \frac{1}{2}\end{align}\]

Chapter 13 Ex.13.2 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’

(ii) E: ‘the card drawn is black’

F: ‘the card drawn is a king’

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

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

Solution

(i) In a deck of \(52\) cards, \(13\) cards are spades and \(4\) cards are aces.

\[\begin{align}&\therefore \;\;{\text{P}}\left( {\text{E}} \right) = {\text{P}}\left( {{\text{the card drawn is a spade}}} \right) = \frac{{13}}{{52}} = \frac{1}{4}\\&\therefore \;\;{\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{the card drawn in an ace}}} \right) = \frac{4}{{52}} = \frac{1}{{13}}\end{align}\]

In the deck of cards, only \(1\) card is an ace of spades.

\[\begin{align} &{\text{P}}\left( {{\text{EF}}} \right) = {\text{P}}\left( {{\text{the card drawn is spade and an ace}}} \right) = \frac{1}{{52}} \hfill \\& {\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = \frac{1}{4} \times \frac{1}{{13}} = \frac{1}{{52}} = {\text{P}}\left( {{\text{EF}}} \right) \hfill \\& \Rightarrow\; \; {\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{EF}}} \right) \hfill \\ \end{align} \]

Therefore, the events E and F are independent.

(ii) In a deck of \(52\) cards, \(26\) cards are black and \(4\) cards are kings.

\[\begin{align}&\therefore \;\;{\text{P}}\left( {\text{E}} \right) = {\text{P}}\left( {{\text{the card drawn is a black}}} \right) = \frac{{26}}{{52}} = \frac{1}{2}\\&\therefore \;\;{\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{the card drawn in an ace}}} \right) = \frac{4}{{52}} = \frac{1}{{13}}\end{align}\]

In the pack of \(52\) cards, \(2\) cards are black as well as kings.

\[\begin{align}{\text{P}}\left( {{\text{EF}}} \right) &= {\text{P}}\left( {{\text{the card drawn is black king}}} \right) = \frac{2}{{52}} = \frac{1}{{26}}\\{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right)& = \frac{1}{2} \times \frac{1}{{13}} = \frac{1}{{26}} = {\text{P}}\left( {{\text{EF}}} \right)\end{align}\]

Therefore, the given events E and F are independent.

(iii) In a deck of \(52\) cards, \(4\) cards are kings, \(4\) cards are queens, and \(4\) cards are jacks.

\[\begin{align}&\therefore \;\;{\text{P}}\left( {\text{E}} \right) = {\text{P}}\left( {{\text{the card drawn is a king or a queen}}} \right) = \frac{8}{{52}} = \frac{2}{{13}}\\&\therefore \;\;{\text{P}}\left( {\text{F}} \right) = {\text{P}}\left( {{\text{the card drawn in a queen or a jack}}} \right) = \frac{8}{{52}} = \frac{2}{{13}}\end{align}\]

There are \(4\) cards which are king or queen and queen or jack.

\[\begin{align}&{\text{P}}\left( {{\text{EF}}} \right) = {\text{P}}\left( {{\text{the card drawn is king or a queen, or queen or a jack}}} \right) = \frac{4}{{52}} = \frac{1}{{13}}\\&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) = \frac{2}{{13}} \times \frac{2}{{13}} = \frac{4}{{169}} \ne \frac{1}{{13}}\\&{\text{P}}\left( {\text{E}} \right) \times {\text{P}}\left( {\text{F}} \right) \ne {\text{P}}\left( {{\text{EF}}} \right)\end{align}\]

Therefore, the given events E and F are not independent.

Chapter 13 Ex.13.2 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.

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

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

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

Solution

Let H denote the students who read Hindi newspaper and E denote the students who read English newspaper.

Given,

\[\begin{align}{\text{P}}\left( {\text{H}} \right) &= 60\% = \frac{{60}}{{100}} = \frac{3}{5}\\{\text{P}}\left( {\text{E}} \right)& = 40\% = \frac{{40}}{{100}} = \frac{2}{5}\\{\text{P}}\left( {{\text{H}} \cap {\text{E}}} \right)& = 20\% = \frac{{20}}{{100}} = \frac{1}{5}\end{align}\]

(i) Probability that a student reads Hindi or English newspaper

\[\begin{align}{\text{P}}\left( {{\text{H}} \cup {\text{E'}}} \right) &= 1 - {\text{P}}\left( {{\text{H}} \cup {\text{E}}} \right)\\&= 1 - \left\{ {{\text{P}}\left( {\text{H}} \right) + {\text{P}}\left( {\text{E}} \right) - {\text{P}}\left( {{\text{H}} \cap {\text{E}}} \right)} \right\}\\& = 1 - \left( {\frac{3}{5} + \frac{2}{5} - \frac{1}{5}} \right)\\&= 1 - \frac{4}{5} = \frac{1}{5}\end{align}\]

(ii) Probability that a randomly chosen student reads English newspaper, if she reads Hindi newspaper, is given by \({\text{P}}\left( {{\text{E}}|{\text{H}}} \right)\).

\({\text{P}}\left( {\left. {\text{E}} \right|{\text{H}}} \right) = \frac{{{\text{P}}\left( {{\text{E}} \cap {\text{H}}} \right)}}{{{\text{P}}\left( {\text{H}} \right)}} = \frac{{\frac{1}{5}}}{{\frac{3}{5}}} = \frac{1}{3}\)

(iii) Probability that a randomly chosen student reads Hindi newspaper, if she reads English newspaper, is given by P (H|E).

\({\text{P}}\left( {\left. {\text{H}} \right|{\text{E}}} \right) = \frac{{{\text{P}}\left( {{\text{H}} \cap {\text{E}}} \right)}}{{{\text{P}}\left( {\text{E}} \right)}} = \frac{{\frac{1}{5}}}{{\frac{2}{5}}} = \frac{1}{2}\)

Chapter 13 Ex.13.2 Question 17

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

\(\begin{align}&\left( {\text{A}} \right)\;0\\\\&\left( {\text{B}} \right)\frac{1}{3}\\\\&\left( {\text{C}} \right)\frac{1}{{12}}\\\\&\left( {\text{D}} \right)\frac{1}{{36}}\end{align}\)

Solution

When two dice are rolled, the number of outcomes is \(36\).

The only even prime number is \(2\).

Let \({\text{E}}\) be the event of getting an even prime number on each die.

\[\begin{align}&\therefore \;\;{\text{E}} = \left\{ {\left( {2,2} \right)} \right\}\\& \Rightarrow\; \; {\text{P}}\left( {\text{E}} \right) = \frac{1}{{36}}\end{align}\]

Therefore, the correct answer is D.

Chapter 13 Ex.13.2 Question 18

Two events A and B will be independent, if

(A) A and B are mutually exclusive

(B)\({\text{P}}\left( {{\text{A' B'}}} \right) = \left[ {1 - {\text{P}}\left( {\text{A}} \right)} \right]\left[ {1 - {\text{P}}\left( {\text{B}} \right)} \right]\)

(C)\({\text{P}}\left( {\text{A}} \right) = {\text{P}}\left( {\text{B}} \right)\)

(D)\({\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) = 1\)

Solution

Two events A and B are said to be independent, if \({\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\)

Let’s take option B.

\[\begin{align}&{\text{P}}\left( {{\text{A' B'}}} \right) = \left[ {1 - {\text{P}}\left( {\text{A}} \right)} \right]\left[ {1 - {\text{P}}\left( {\text{B}} \right)} \right]\\& \Rightarrow\; \; {\text{P}}\left( {{\text{A'}} \cap {\text{B'}}} \right) = 1 - {\text{P}}\left( {\text{A}} \right) - {\text{P}}\left( {\text{B}} \right) + {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &\Rightarrow\; \; 1 - {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = 1 - {\text{P}}\left( {\text{A}} \right) - {\text{P}}\left( {\text{B}} \right) + {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &\Rightarrow\; \; {\text{P}}\left( {{\text{A}} \cup {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right) - {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\& \Rightarrow\; \; {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right){\text{ + P}}\left( {\text{B}} \right){\text{ - P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\& \Rightarrow\; \; {\text{P}}\left( {{\text{AB}}} \right) = {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\end{align}\]

This implies that A and B are independent, if \({\text{ P}}\left( {{\text{A' B'}}} \right) = \left[ {1 - {\text{P}}\left( {\text{A}} \right)} \right]\left[ {1 - {\text{P}}\left( {\text{B}} \right)} \right]\)

A.\({\text{Let}}\;{\text{P}}\left( {\text{A}} \right) = {\text{m}},\;{\text{P}}\left( {\text{B}} \right) = {\text{n, 0}} < {\text{m}},{\text{n}} < {\text{1}}\)

A and B are mutually exclusive.

\[\begin{align}&\therefore \;\;{\text{A}} \cap {\text{B}} = \phi \\& \Rightarrow\; \; {\text{P}}\left( {{\text{AB}}} \right) = 0\\&{\text{However, P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = {\text{mn}} \ne 0\\&\therefore \;{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) \ne {\text{P}}\left( {{\text{AB}}} \right)\end{align}\]

C.Let A: Event of getting an odd number on throw of a die\( = \left\{ {1,3,5} \right\}\)

\( \Rightarrow\; \; {\text{P}}\left( {\text{A}} \right) = \frac{3}{6} = \frac{1}{2}\)

Let B: Event of getting an even number on throw of a die\( = \left\{ {2,4,6} \right\}\)

\[\begin{align} &\Rightarrow\; \; {\text{P}}\left( {\text{B}} \right) = \frac{3}{6} = \frac{1}{2}\\&{\text{Here,}}\;{\text{A}} \cap {\text{B}} = \phi \\&\therefore \;\;{\text{P}}\left( {{\text{AB}}} \right) = 0\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) = \frac{1}{4} \ne 0\\&{\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right) \ne {\text{P}}\left( {{\text{AB}}} \right)\end{align}\]

D. From the above example, it can be seen that,

\({\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) = \frac{1}{2} + \frac{1}{2} = 1\)

However, it cannot be inferred that A and B are independent.

Thus, the correct answer is B.

  
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