# Miscellaneous Exercise Probability Solution - NCERT Class 12

## Chapter 13 Ex.13.ME Question 1

A and B are two events such that $${\rm{P}}\left( {\rm{A}} \right) \ne 0$$, find $${\text{ P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right)$$, if

(i) A is a subset of B

(ii)$${\text{A}} \cap {\text{B}} = \phi$$.

### Solution

Given,$${\text{P}}\left( {\text{A}} \right) \ne 0$$

(i) A is a subset of B.

$$\Rightarrow \; {\text{A}} \cap {\text{B}} = {\text{A}}$$

\begin{align}&\therefore \;\;{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {{\text{B}} \cap {\text{A}}} \right) = {\text{P}}\left( {\text{A}} \right)\\&\therefore \;\;{\text{P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right) = \frac{{{\text{P}}\left( {{\text{B}} \cap {\text{A}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}} = \frac{{{\text{P}}\left( {\text{A}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}} = 1\end{align}

\begin{align}{\text{(ii) A}} \cap {\text{B}} = \phi \\&\Rightarrow \; \;{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = 0\\&\;\therefore \;\;{\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)}} = 0\end{align}

## Chapter 13 Ex.13.ME Question 2

A couple has two children,

(i) Find the probability that both children are males, if it is known that at least one of the children is male.

(ii) Find the probability that both children are females, if it is known that the elder child is a female.

### Solution

If a couple has two children, then the sample space is $${\text{S}} = \left\{ {\left( {{\text{B, B}}} \right){\text{,}}\;\left( {{\text{B, G}}} \right){\text{, }}\left( {{\text{G, B}}} \right){\text{, }}\left( {{\text{G, G}}} \right)} \right\}$$

(i) Let E and F respectively denote the events that both children are male and atleast one children is a male.

\begin{align}{\text{E}} \cap {\text{F}} &= \left\{ {\left( {{\text{G,G}}} \right)} \right\} \Rightarrow \; {\text{P}}\left( {{\text{E}} \cap {\text{F}}} \right) &= \frac{1}{4}\\{\text{P}}\left( {\text{E}} \right) &= \frac{1}{4}\\{\text{P}}\left( {\text{F}} \right) &= \frac{3}{4}\end{align}

\begin{align} \Rightarrow \; {\text{P}}\left( {\left. {\text{E}} \right|{\text{F}}} \right) &= \frac{{{\text{P}}\left( {{\text{E}} \cap {\text{F}}} \right)}}{{{\text{P}}\left( {\text{F}} \right)}}\\ &= \frac{{\frac{1}{4}}}{{\frac{3}{4}}} = \frac{1}{3}\end{align}

(ii) Let C and D respectively denote the events that both children are females and the elder child is a female.

\begin{align}{\text{C}} &= \left\{ {\left( {{\text{G,G}}} \right)} \right\} \Rightarrow \; {\text{P}}\left( {\text{C}} \right) = \frac{1}{4}\\{\text{D}} &= \left\{ {\left( {{\text{G,B}}} \right),\left( {{\text{G,G}}} \right)} \right\} \Rightarrow \; {\text{P}}\left( {\text{D}} \right) = \frac{2}{4}\\{\text{C}} \cap {\text{D}}& = \left\{ {\left( {{\text{G,G}}} \right)} \right\} \Rightarrow \; {\text{P}}\left( {{\text{C}} \cap {\text{D}}} \right) = \frac{1}{4}\\\therefore& \;\;{\text{P}}\left( {\left. {\text{C}} \right|{\text{D}}} \right) = \frac{{{\text{P}}\left( {{\text{C}} \cap {\text{D}}} \right)}}{{{\text{P}}\left( {\text{D}} \right)}} = \frac{{\frac{1}{4}}}{{\frac{2}{4}}} = \frac{1}{2}\end{align}

## Chapter 13 Ex.13.ME Question 3

Suppose that $$5\%$$ of men and $$0.25\%$$ of women have grey hair. A haired person is selected at random. What is the probability of this person being male?

Assume that there are equal numbers of males and females.

### Solution

Given, $$5\%$$ of men and $$0.25\%$$ of women have grey hair.

Thus, percentage of people with grey hair $$= \left( {5 + 0.25} \right)\% = 5.25\%$$

Probability that the selected haired person is a male$$= \frac{5}{{5.25}} = \frac{{20}}{{21}}$$

## Chapter 13 Ex.13.ME Question 4

Suppose that 90% of people are right-handed. What is the probability that at most 6 of a random sample of 10 people are right-handed?

### Solution

A person can be either right-handed or left-handed.

Given, 90% of the people are right-handed.

\begin{align}\therefore \;\;p& = {\text{P}}\left( {{\text{right - handed}}} \right) = \;\frac{9}{{10}}\\\therefore \;\;q &= {\text{P}}\left( {{\text{left - handed}}} \right) = 1 - p = 1 - \frac{9}{{10}} = \frac{1}{{10}}\end{align}

Using binominal distribution, the probability that more than 6 people are right-handed is given by

$$\sum\limits_{r = 7}^{10} {{}^{10}{{\text{C}}_r}{p^r}{q^{n - r}}} = \sum\limits_{r = 7}^{10} {{}^{10}{{\text{C}}_r}{{\left( {\frac{9}{{10}}} \right)}^r} \times {{\left( {\frac{1}{{10}}} \right)}^{10 - r}}}$$

Therefore, the probability that at most 6 people are right-handed

\begin{align}&= 1 - {\text{P}}\left( {{\text{more than 6 are right - handed}}} \right)\\&= 1 - \sum\limits_{r = 7}^{10} {{}^{10}{{\text{C}}_r}{{\left( {0.9} \right)}^r} \times {{\left( {0.1} \right)}^{10 - r}}} \end{align}

## Chapter 13 Ex.13.ME Question 5

An urn contains $$25$$ balls of which $$10$$ balls bear a mark ‘X’ and the remaining $$15$$ bear a mark ‘Y’.A ball is drawn at random from the urn, its mark is noted down, and it is replaced. If 6 balls are drawn in this way, find the probability that

(i) all will bear ‘X’ mark.

(ii) not more than $$2$$ will bear ‘Y’ mark.

(iii) at least one ball will bear ‘Y’ mark.

(iv) the number of balls with ‘X’ mark and ‘Y’ mark will be equal.

### Solution

Total number of balls in the urn$${\text{ = 25}}$$

Balls bearing mark ‘X’ $${\text{ = 10}}$$

Balls bearing mark ‘Y’ $${\text{ = 15}}$$

\begin{align}p& = {\text{P}}\left( {{\text{ball bearing mark 'X'}}} \right) = \frac{{10}}{{25}} = \frac{2}{5}\\q &= {\text{P}}\left( {{\text{Balls bearing mark 'Y'}}} \right) = \frac{{15}}{{25}} = \frac{3}{5}\end{align}

Let Z be the random variable that represents the number of balls with ‘Y’ mark on them in the trials.

Clearly, Z has a binomial distribution with $$n = 6{\text{ and }}p = \frac{2}{5}$$.

$$\therefore {\text{ P}}\left( {{\text{Z}} = z} \right) = {}^n{{\text{C}}_z}{p^{n - z}}{q^z}$$

(i)$${\text{P}}\left( {{\text{all will bear 'X' mark}}} \right) = {\text{P}}\left( {{\text{Z}} = 0} \right) = {}^6{{\text{C}}_0}{\left( {\frac{2}{5}} \right)^6} = {\left( {\frac{2}{5}} \right)^6}$$

(ii)$${\text{P}}\left( {{\text{not more than 2 bear 'Y' mark}}} \right) = {\text{P}}\left( {{\text{Z}} \le 2} \right)$$

\begin{align}&= {\text{P}}\left( {{\text{Z}} = 0} \right) + {\text{P}}\left( {{\text{Z}} = 1} \right) + {\text{P}}\left( {{\text{Z}} = 2} \right)\\&= {}^6{{\text{C}}_0} \times {\left( p \right)^6} \times {\left( q \right)^0} + {}^6{{\text{C}}_1} \times {\left( p \right)^5} \times {\left( q \right)^1} + {}^6{{\text{C}}_2} \times {\left( p \right)^4} \times {\left( q \right)^2}\\&= {\left( {\frac{2}{5}} \right)^6} + 6 \times {\left( {\frac{2}{5}} \right)^5} \times \left( {\frac{3}{5}} \right) + 15 \times {\left( {\frac{2}{5}} \right)^4} \times {\left( {\frac{3}{5}} \right)^2}\end{align}

\begin{align}&= {\left( {\frac{2}{5}} \right)^4} \times \left[ {{{\left( {\frac{2}{5}} \right)}^2} + 6 \times \left( {\frac{2}{5}} \right) \times \left( {\frac{3}{5}} \right) + 15 \times {{\left( {\frac{3}{5}} \right)}^2}} \right]\\&= {\left( {\frac{2}{5}} \right)^4} \times \left[ {\frac{4}{{25}} + \frac{{36}}{{25}} + \frac{{135}}{{25}}} \right]\\&= {\left( {\frac{2}{5}} \right)^4} \times \left[ {\frac{{175}}{{25}}} \right]\\&= 7 \times {\left( {\frac{2}{5}} \right)^4} = 0.1792\end{align}

(iii)$${\text{P}}\left( {{\text{at least one ball bear 'Y' mark}}} \right) = {\text{P}}\left( {{\text{Z}} \ge 1} \right) = 1 - {\text{P}}\left( {{\text{Z}} = 0} \right)$$

$$= 1 - {\left( {\frac{2}{5}} \right)^6}$$

(iv)$${\text{P}}\left( {{\text{equal number of balls with 'X' mark}}\;{\text{and 'Y' mark}}} \right) = {\text{P}}\left( {{\text{Z}} = 3} \right)$$

\begin{align}&= {}^6{{\text{C}}_3} \times {\left( {\frac{2}{5}} \right)^3} \times {\left( {\frac{3}{5}} \right)^3}\\&= \frac{{20 \times 8 \times 27}}{{15625}}\\&= \frac{{864}}{{3125}}\end{align}

## Chapter 13 Ex.13.ME Question 6

In a hurdle race, a player has to cross $$10$$ hurdles. The probability that he will clear each hurdle is $$\frac{5}{6}$$. What is the probability that he will knock down fewer than $$2$$ hurdles?

### Solution

Let p and q respectively be the probability that the player will clear and knock down the hurdle.

\begin{align}&\therefore \;\;p = \frac{5}{6}\\&\Rightarrow \; q = 1 - p = 1 - \frac{5}{6} = \frac{1}{6}\end{align}

Let X be the random variable that represents the number of times the player will knock down the hurdle.

Thus, by binomial distribution, we get

$$\therefore {\text{ P}}\left( {{\text{X}} = x} \right) = {}^n{{\text{C}}_x}{p^{n - x}}{q^x}$$

\begin{align}{\text{P}}\left( {{\text{Player knocking down less than 2 hurdles}}} \right)& = {\text{P}}\left( {{\text{X}} < 2} \right)\\&= {\text{P}}\left( {{\text{X}} = 0} \right) + {\text{P}}\left( {{\text{X}} = 1} \right)\\&= {}^{10}{{\text{C}}_0} \times {\left( q \right)^0} \times {\left( p \right)^{10}} + {}^{10}{{\text{C}}_1} \times {\left( q \right)^1} \times {\left( p \right)^9}\\&= {\left( {\frac{5}{6}} \right)^{10}} + 10 \times \frac{1}{6} \times {\left( {\frac{5}{6}} \right)^9}\\&= {\left( {\frac{5}{6}} \right)^9} + \left[ {\frac{5}{6} + \frac{{10}}{6}} \right]\\&= \frac{5}{2} \times {\left( {\frac{5}{6}} \right)^9}\\&= \frac{{{{\left( 5 \right)}^{10}}}}{{2 \times {{\left( 6 \right)}^9}}}\end{align}

## Chapter 13 Ex.13.ME Question 7

A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

### Solution

The probability of getting a six in a throw of die is $$\frac{1}{6}$$ and not getting a six is $$\frac{5}{6}$$.

Let $$p = \frac{1}{6}{\text{ and }}q = \frac{5}{6}$$

The probability that the $$2$$ sixes come in the first five throws of the die is

$${}^5{{\text{C}}_2} \times {\left( {\frac{1}{6}} \right)^2} \times {\left( {\frac{5}{6}} \right)^3} = \frac{{10 \times {{\left( 5 \right)}^3}}}{{{{\left( 6 \right)}^5}}}$$

Probability that third six comes in the sixth throw$$= \frac{{10 \times {{\left( 5 \right)}^3}}}{{{{\left( 6 \right)}^5}}} \times \frac{1}{6}$$

\begin{align}&= \frac{{10 \times 125}}{{{{\left( 6 \right)}^6}}}\\&= \frac{{10 \times 125}}{{46656}}\\&= \frac{{625}}{{23328}}\end{align}

## Chapter 13 Ex.13.ME Question 8

If a leap year is selected at random, what is the change that it will contain $$53$$ Tuesdays?

### Solution

In a leap year, there are $$366$$ days i.e.,$$52$$weeks and $$2$$ days.

In $$52$$ weeks, there are $$52$$ Tuesdays.

Therefore, the probability that the leap year will contain $$53$$ Tuesdays is equal to the probability that the remaining $$2$$ days will be Tuesdays.

The remaining $$2$$ days can be any of the following:

Monday and Tuesday, Tuesday and Wednesday, Wednesday and Thursday, Thursday and Friday, Friday and Saturday, Saturday and Sunday and Sunday and Monday.

Total number of cases $$= 7$$

Favourable cases $$= 2$$

Probability that a leap year will have $$53$$ Tuesdays$$= \frac{2}{7}$$

## Chapter 13 Ex.13.ME Question 9

An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be at least $$4$$ successes.

### Solution

Given, the probability of success is twice the probability of failure.

Let the probability of failure be $$x$$.

Probability of success $$= 2x$$

\begin{align}&x + 2x = 1\\&\Rightarrow \; 3x = 1\\&\Rightarrow \; x = \frac{1}{3}\\&\therefore \;2x = \frac{2}{3}\end{align}

Let $$p = \frac{1}{3}$$and $$q = \frac{2}{3}$$

Let X be the random variable that represents the number of successes in six trials.

By binomial distribution, we obtain

$${\text{P}}\left( {{\text{X}} = x} \right) = {}^n{{\text{C}}_x}{p^{n - x}}{q^x}$$

\begin{align}{\text{Probability of at least 4 successes}}& = {\text{P}}\left( {{\text{X}} \ge 4} \right)\\&= {\text{P}}\left( {{\text{X}} = 4} \right) + {\text{P}}\left( {{\text{X}} = 5} \right) + {\text{P}}\left( {{\text{X}} = 6} \right)\\&= {}^6{{\text{C}}_4} \times {\left( {\frac{2}{3}} \right)^4} \times {\left( {\frac{1}{3}} \right)^2} + {}^6{{\text{C}}_5} \times {\left( {\frac{2}{3}} \right)^5} \times \left( {\frac{1}{3}} \right) + {}^6{{\text{C}}_6} \times {\left( {\frac{2}{3}} \right)^6}\\&= \frac{{15 \times {{\left( 2 \right)}^4}}}{{{3^6}}} + \frac{{6 \times {{\left( 2 \right)}^5}}}{{{3^6}}} + \frac{{{{\left( 2 \right)}^6}}}{{{3^6}}}\\&= \frac{{{{\left( 2 \right)}^4}}}{{{{\left( 3 \right)}^6}}} \times \left[ {15 + 12 + 4} \right]\\&= \frac{{31 \times {{\left( 2 \right)}^4}}}{{{{\left( 3 \right)}^6}}}\\&= \frac{{31}}{9} \times {\left( {\frac{2}{3}} \right)^4}\end{align}

## Chapter 13 Ex.13.ME Question 10

How many times must a man toss a fair coin so that the probability of having at least one head is more than $$90\%$$?

### Solution

Let the man toss the coin n times. The n tosses are n Bernoulli trials.

Probability$$\left( p \right)$$of getting a head at the toss of a coin is $$\frac{1}{2}$$

\begin{align}p &= \frac{1}{2}{\text{, }}q = \frac{1}{2}\\{\text{P}}\left( {{\text{X}} = x} \right) &= {}^n{{\text{C}}_x}{p^{n - x}}{q^x} = {}^n{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^{n - x}}{\left( {\frac{1}{2}} \right)^x} = {}^n{{\text{C}}_x}{\left( {\frac{1}{2}} \right)^n}\end{align}

\begin{align}&{\text{P}}\left( {{\text{getting at least one head}}} \right) > \frac{{90}}{{100}}\\&{\text{P}}\left( {x \ge 1} \right) > 0.9\\&1 - {\text{P}}\left( {x = 0} \right) > 0.9\\&1 - {}^n{{\text{C}}_0} \times \frac{1}{{{2^n}}} > 0.9\\&{}^n{{\text{C}}_0} \times \frac{1}{{{2^n}}} < 0.1\\&\frac{1}{{{2^n}}} < 0.1\\&{2^n} > \frac{1}{{0.1}}\\&{2^n} > 10\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;...\left( 1 \right)\end{align}

The minimum value of n that satisfies the given inequality is $$4$$.

Thus, the man should toss the coin $$4$$ or more than $$4$$ times.

## Chapter 13 Ex.13.ME Question 11

In a game, a man wins a rupee for a six and loss a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.

### Solution

Here, the probability of getting a six is$$\frac{1}{6}$$and the probability of not getting a $$6$$ is$$\frac{5}{6}$$.

Three cases can occur.

(a) If he gets a six in the first throw, then the required probability is$$\frac{1}{6}$$

Amount he will receive$$= {\text{Re 1}}$$.

(b) If he does not get a six in the first throw and gets a six in the second throw, then probability$$= \left( {\frac{5}{6} \times \frac{1}{6}} \right) = \frac{5}{{36}}$$

Amount he will receive$$= - {\text{Re 1}} + {\text{Re 1}} = 0$$

(c) If he does not get a six in the first two throws and gets a six in the third throw, then probability 

$${\text{Amount he will receive}} = - {\text{Re 1}} - {\text{Re 1}} + {\text{Re 1}} = - 1$$

Probability that he does not get a six in any of the three throws $$= \frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} = \frac{{125}}{{216}}$$

Expected value he can win$$= \frac{1}{6} \times \left( 1 \right) + \left( {\frac{5}{6} \times \frac{1}{6}} \right) \times \left( 0 \right) + \left( {{{\left( {\frac{5}{6}} \right)}^2} \times \frac{1}{6}} \right) \times \left( { - 1} \right)$$

\begin{align}&= \frac{1}{6} - \frac{{25}}{{216}}\\&= \frac{{36 - 25}}{{216}} = \frac{{11}}{{216}}\end{align}

## Chapter 13 Ex.13.ME Question 12

Suppose we have four boxes. A, B, C and D containing coloured marbles as given below.

 Box Marble Colour Red White Black A $$1$$ $$6$$ $$3$$ B $$6$$ $$2$$ $$2$$ C $$8$$ $$1$$ $$1$$ D $$0$$ $$6$$ $$4$$

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A?, box B?, box C?

### Solution

Let R be the event of drawing the red marble.

Let $${{\text{E}}_{\text{A}}}{\text{, }}{{\text{E}}_{\text{B}}}{\text{ and }}{{\text{E}}_{\text{C}}}$$ respectively denote the events of selecting the box A, B, and C.

Total number of marbles $$= 40$$

Number of red marbles $$= 15$$

$$\therefore \;\;{\text{P}}\left( {\text{R}} \right) = \frac{{15}}{{40}} = \frac{3}{8}$$

Probability of drawing the red marble from box A is given by $${\text{P}}\left( {\left. {{\text{E}}{}_{\text{A}}} \right|{\text{R}}} \right)$$.

$$\therefore \;\;{\text{P}}\left( {\left. {{\text{E}}{}_{\text{A}}} \right|{\text{R}}} \right) = \frac{{{\text{P}}\left( {{\text{E}}{}_{\text{A}} \cap {\text{R}}} \right)}}{{{\text{P}}\left( {\text{R}} \right)}} = \frac{{\frac{1}{{40}}}}{{\frac{3}{8}}} = \frac{1}{{15}}$$

Probability that the red marble is from box B is $${\text{P}}\left( {\left. {{\text{E}}{}_{\text{B}}} \right|{\text{R}}} \right)$$.

$$\therefore \;\;{\text{P}}\left( {\left. {{\text{E}}{}_{\text{B}}} \right|{\text{R}}} \right) = \frac{{{\text{P}}\left( {{\text{E}}{}_{\text{B}} \cap {\text{R}}} \right)}}{{{\text{P}}\left( {\text{R}} \right)}} = \frac{{\frac{6}{{40}}}}{{\frac{3}{8}}} = \frac{2}{5}$$

Probability that the red marble is from box C is $${\text{P}}\left( {\left. {{\text{E}}{}_{\text{C}}} \right|{\text{R}}} \right)$$.

$$\therefore \;\;{\text{P}}\left( {\left. {{\text{E}}{}_{\text{C}}} \right|{\text{R}}} \right) = \frac{{{\text{P}}\left( {{\text{E}}{}_{\text{C}} \cap {\text{R}}} \right)}}{{{\text{P}}\left( {\text{R}} \right)}} = \frac{{\frac{8}{{40}}}}{{\frac{3}{8}}} = \frac{8}{{15}}$$

## Chapter 13 Ex.13.ME Question 13

Assume that the changes of the patient having a heart attack are $$40\%$$. It is also assumed that a meditation and yoga course reduce the risk of heart attack by $$30 \%$$ and prescription of certain drug reduces its changes by $$25\%$$. At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that the patient followed a course of meditation and yoga?

### Solution

Let $${\text{A, }}{{\text{E}}_1}{\text{ and }}{{\text{E}}_2}$$ respectively denote the events that a person has a heart attack, the selected person followed the course of yoga and meditation, and the person adopted the drug prescription.

$$\therefore \;\;{\text{P}}\left( {\text{A}} \right) = 0.40$$

$${\text{P}}\left( {{{\text{E}}_1}} \right) = {\text{P}}\left( {{{\text{E}}_2}} \right) = \frac{1}{2}$$

\begin{align}&\Rightarrow \; \;{\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_1}} \right) = 0.40 \times 0.70 = 0.28\\&\Rightarrow \; \;{\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_2}} \right) = 0.40 \times 0.75 = 0.30\end{align}

Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by $${\text{P}}\left( {\left. {{{\text{E}}_1}} \right|{\text{A}}} \right)$$.

\begin{align}{\text{P}}\left( {\left. {{{\text{E}}_1}} \right|{\text{A}}} \right) &= \frac{{{\text{P}}\left( {{{\text{E}}_1}} \right) \times {\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_1}} \right)}}{{{\text{P}}\left( {{{\text{E}}_1}} \right) \times {\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_1}} \right) + {\text{P}}\left( {{{\text{E}}_2}} \right) \times {\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_2}} \right)}}\\&= \frac{{\frac{1}{2} \times 0.28}}{{\frac{1}{2} \times 0.28 + \frac{1}{2} \times 0.30}} = \frac{{14}}{{29}}\end{align}

## Chapter 13 Ex.13.ME Question 14

If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $$\frac{1}{2}$$).

### Solution

The total number of determinants of second order with each element being $$0$$ or $$1$$ is $${\left( 2 \right)^4} = 16$$.

The value of determinant is positive in the following cases.

$$\left| \begin{array}{l}1\;\;\;\;0\\0\;\;\;\;1\end{array} \right|,\;\left| \begin{array}{l}1\;\;\;\;1\\0\;\;\;\;1\end{array} \right|,\;\left| \begin{array}{l}1\;\;\;\;0\\1\;\;\;\;1\end{array} \right|$$

Therefore, Required probability$$= \frac{3}{{16}}$$

## Chapter 13 Ex.13.ME Question 15

An electronic assembly consists of two subsystems, say, $$A$$ and $$B$$. From previous testing procedures, the following probabilities are assumed to be known:

\begin{align}&{\text{P}}\left( {{\text{A fails}}} \right) = 0.2\\&{\text{P}}\left( {{\text{B fails alone}}} \right) = 0.15\\&{\text{P}}\left( {{\text{A and B fails}}} \right) = 0.15\end{align}

Evaluate the following probabilities

\begin{align}&\left( {\text{i}} \right)\;{\text{P}}\left( {\left. {{\text{A fails}}} \right|{\text{B has failed}}} \right)\\\\&\left( {{\text{ii}}} \right)\;{\text{P}}\left( {{\text{A fails alone}}} \right) \end{align}

### Solution

Let the event in which $$A$$ fails and $$B$$ fails be denote by $${\text{E}}{}_{\text{A}}{\text{ and E}}{}_{\text{B}}$$

\begin{align}&{\text{P}}\left( {{\text{E}}{}_{\text{A}}} \right) = 0.2\\&{\text{P}}\left( {{\text{E}}{}_{\text{A}}{\text{ }} \cap {\text{ E}}{}_{\text{B}}} \right) = 0.15\end{align}

\begin{align}{\text{P}}\left( {{\text{B fails alone}}} \right)& = {\text{P}}\left( {{\text{E}}{}_{\text{B}}} \right) - {\text{P}}\left( {{\text{E}}{}_{\text{A}} \cap {\text{ E}}{}_{\text{B}}} \right)\\\therefore \;0.15 &= {\text{P}}\left( {{\text{E}}{}_{\text{B}}} \right) - 0.15\\\therefore \;{\text{P}}\left( {{\text{E}}{}_{\text{B}}} \right)& = 0.3\end{align}

$$\left( {\text{i}} \right)\;\;{\text{P}}\left( {\left. {{\text{E}}{}_{\text{A}}} \right|{\text{E}}{}_{\text{B}}} \right) = \frac{{{\text{P}}\left( {{\text{E}}{}_{\text{A}} \cap {\text{E}}{}_{\text{B}}} \right)}}{{{\text{P}}\left( {{\text{E}}{}_{\text{B}}} \right)}} = \frac{{0.15}}{{0.3}} = 0.5$$

\begin{align}\left( {{\text{ii}}} \right)\;\;{\text{P}}\left( {{\text{A fails alone}}} \right) &= {\text{P}}\left( {{\text{E}}{}_{\text{A}}} \right) - {\text{P}}\left( {{\text{E}}{}_{\text{A}}{\text{ and E}}{}_{\text{B}}} \right)\\&= 0.2 - 0.15 = 0.05\end{align}

## Chapter 13 Ex.13.ME Question 16

Bag I contains $$3$$ red and $$4$$ black balls and Bag II contains $$4$$ red and $$​​5$$ black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

### Solution

Let $${\text{E}}{}_1{\text{ and E}}{}_2$$ respectively denote the event that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.

$${\text{P}}\left( {{\text{E}}{}_1} \right) = \frac{3}{7}{\text{ and P}}\left( {{\text{E}}{}_2} \right) = \frac{4}{7}$$

Let A be the event that the ball drawn is red.

When a red ball is transferred from bag I to II,

$${\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_1}} \right) = \frac{5}{{10}} = \frac{1}{2}$$

When a black ball is transferred from bag I to II,

$${\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_2}} \right) = \frac{4}{{10}} = \frac{2}{5}$$

\begin{align}{\text{P}}\left( {\left. {{{\text{E}}_2}} \right|{\text{A}}} \right) &= \frac{{{\text{P}}\left( {{{\text{E}}_2}} \right) \times {\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_2}} \right)}}{{{\text{P}}\left( {{{\text{E}}_1}} \right) \times {\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_1}} \right) + {\text{P}}\left( {{{\text{E}}_2}} \right) \times {\text{P}}\left( {\left. {\text{A}} \right|{{\text{E}}_2}} \right)}}\\&= \frac{{\frac{4}{7} \times \frac{2}{5}}}{{\frac{3}{7} \times \frac{1}{2} + \frac{4}{7} \times \frac{2}{5}}} = \frac{{16}}{{31}}\end{align}

## Chapter 13 Ex.13.ME Question 17

If A and B are two events such that $$P (A) ≠ 0$$ and $$P (B|A) = 1$$, then

\begin{align}&{\text{(A) A}} \subset {\text{B}}\\&{\text{(B) B}} \subset {\text{A}}\\&{\text{(C) B}} = \phi \\&{\text{(D) A}} = \phi \end{align}

### Solution

Given,$${\text{P}}\left( {\text{A}} \right) \ne {\text{0 and P}}\left( {\left. {\text{A}} \right|{\text{B}}} \right) = 1$$.

\begin{align}\therefore \;\;{\text{P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right) &= \frac{{{\text{P}}\left( {{\text{B}} \cap {\text{A}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}\\1 &= \frac{{{\text{P}}\left( {{\text{B}} \cap {\text{A}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}}\\{\text{P}}\left( {\text{A}} \right) &= {\text{P}}\left( {{\text{B}} \cap {\text{A}}} \right)\\ \Rightarrow \; {\text{A}} \subset {\text{B}}\end{align}

Thus, the correct answer is $$A$$.

## Chapter 13 Ex.13.ME Question 18

If and $${\text{P}}\left( {\left. {\text{A}} \right|{\text{B}}} \right) > {\text{P}}\left( {\text{A}} \right)$$, then which of the following is correct:

A. $${\rm{P}}\left( {\left. {\rm{B}} \right|{\rm{A}}} \right) < {\rm{P}}\left( {\rm{B}} \right)$$

B. $${\rm{P}}\left( {{\rm{A}} \cap {\rm{B}}} \right) < {\rm{P}}\left( {\rm{A}} \right) \times {\rm{P}}\left( {\rm{B}} \right)$$

C. $${\rm{P}}\left( {\left. {\rm{B}} \right|{\rm{A}}} \right) > {\rm{P}}\left( {\rm{B}} \right)$$

D. $${\rm{P}}\left( {\left. {\rm{B}} \right|{\rm{A}}} \right) = {\rm{P}}\left( {\rm{B}} \right)$$

### Solution

Given, $${\text{P}}\left( {\left. {\text{A}} \right|{\text{B}}} \right) > {\text{P}}\left( {\text{A}} \right)$$

\begin{align} &\Rightarrow \; \;\;\frac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}} > {\text{P}}\left( {\text{A}} \right)\\ &\Rightarrow \; \;\;{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) > {\text{P}}\left( {\text{A}} \right) \times {\text{P}}\left( {\text{B}} \right)\\ &\Rightarrow \; \;\;\frac{{{\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right)}}{{{\text{P}}\left( {\text{A}} \right)}} > {\text{P}}\left( {\text{B}} \right)\\ &\Rightarrow \; {\text{P}}\left( {\left. {\text{B}} \right|{\text{A}}} \right) > {\text{P}}\left( {\text{B}} \right)\end{align}

Thus, the correct answer is C.

## Chapter 13 Ex.13.ME Question 19

If A and B are any two events such that $${\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A and B}}} \right) = {\text{P}}\left( {\text{A}} \right)$$, then

A. $${\rm{P}}\left( {\left. {\rm{B}} \right|{\rm{A}}} \right) = 1$$

B. $${\rm{P}}\left( {\left. {\rm{A}} \right|{\rm{B}}} \right) = 1$$

C. $${\rm{P}}\left( {\left. {\rm{B}} \right|{\rm{A}}} \right) = 0$$

D. $${\rm{P}}\left( {\left. {\rm{A}} \right|{\rm{B}}} \right) = {\rm{0}}$$

### Solution

Given, $${\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A and B}}} \right) = {\text{P}}\left( {\text{A}} \right)$$

\begin{align}&\Rightarrow \; {\text{P}}\left( {\text{A}} \right) + {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{A}} \right)\\&\Rightarrow \; {\text{P}}\left( {\text{B}} \right) - {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = 0\\& \Rightarrow \; {\text{P}}\left( {{\text{A}} \cap {\text{B}}} \right) = {\text{P}}\left( {\text{B}} \right)\\&\therefore \;{\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)}} = \frac{{{\text{P}}\left( {\text{B}} \right)}}{{{\text{P}}\left( {\text{B}} \right)}} = 1\end{align}

Thus, the correct answer is B.

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