## Chapter 13 Probability Miscellaneous Exercise

### Miscellaneous Exercise on Chapter 13

**1.** $A$ and $B$ are two events such that $P(A) \neq 0$. Find $P(B \mid A)$, if

(i) $A$ is a subset of $B$

(ii) $A \cap B=\phi$

## Show Answer

**Solution**

It is given that, $P(A) \neq 0$

(i) $A$ is a subset of $B$.

$\Rightarrow A \cap B=A$

$\therefore P(A \cap B)=P(B \cap A)=P(A)$

$\therefore P(B \mid A)=\frac{P(B \cap A)}{P(A)}=\frac{P(A)}{P(A)}=1$

(ii) $A \cap B=\phi$

$\Rightarrow P(A \cap B)=0$

$\therefore P(B \mid A)=\frac{P(A \cap B)}{P(A)}=0$

**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.

## Show Answer

**Solution**

If a couple has two children, then the sample space is

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

(i) Let $E$ and $F$ respectively denote the events that both children are males and at least one of the children is a male. $\therefore E \cap F=\{(b, b)\} \Rightarrow P(E \cap F)=\frac{1}{4}$

$P(E)=\frac{1}{4}$

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

$\Rightarrow P(E \mid F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

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

$ \begin{aligned} & A=\{(g, g)\} \Rightarrow P(A)=\frac{1}{4} \\ & B=\{(g, b),(g, g)\} \Rightarrow P(B)=\frac{2}{4} \\ & A \cap B=\{(g, g)\} \Rightarrow P(A \cap B)=\frac{1}{4} \\ & P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{2}}{\frac{2}{4}}=\frac{1}{2} \end{aligned} $

**3.** Suppose that $5 \%$ of men and $0.25 \%$ of women have grey hair. A grey haired person is selected at random. What is the probability of this person being male? Assume that there are equal number of males and females.

## Show Answer

**Solution**

It is given that $5 \%$ of men and $0.25 \%$ of women have grey hair.

Therefore, percentage of people with grey hair $=(5+0.25) \%=5.25 \%$

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

**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?

## Show Answer

**Solution**

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

It is given that $90 \%$ of the people are right-handed.

$\therefore p=P($ right-handed $)=\frac{9}{10}$

$q=P($ left-handed $)=1-\frac{9}{10}=\frac{1}{10}$

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

$\sum _{r=7}^{10}{ }^{10} C_r p^{r} q^{n-r}=\sum _{r=7}^{10}{ }^{10} C_r(\frac{9}{10})^{r}(\frac{1}{10})^{10-r}$

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

$=1-P$ (more than 6 are right-handed)

$=1-\sum _{r=7}^{10}{ }^{10} C_r(0.9)^{r}(0.1)^{10-r}$

**5.** If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

## Show Answer

**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

Monday and Tuesday

Tuesday and Wednesday

Wednesday and Thursday

Thursday and Friday

Friday and Saturday

Saturday and Sunday

Sunday and Monday

Total number of cases $=7$

Favourable cases $=2$

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

**6.** Suppose we have four boxes A,B,C and D containing coloured marbles as given below: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 $C$ ?

Box | Marble colour | ||
---|---|---|---|

Red | White | Black | |

A | 1 | 6 | 3 |

B | 6 | 2 | 2 |

C | 8 | 1 | 1 |

D | 0 | 6 | 4 |

## Show Answer

**Solution**

Let $R$ be the event of drawing the red marble.

Let $E_A, E_B$, and $E_C$ respectively denote the events of selecting the box $A, B$, and $C$.

Total number of marbles $=40$

Number of red marbles $=15$

$\therefore P(R)=\frac{15}{40}=\frac{3}{8}$

Probability of drawing the red marble from box $A$ is given by $P(E_A \mid R)$.

$\therefore P(E_A \mid R)=\frac{P(E_A \cap R)}{P(R)}=\frac{\frac{1}{40}}{\frac{3}{8}}=\frac{1}{15}$

Probability that the red marble is from box $B$ is $P(E_B \mid R)$.

$\Rightarrow P(E_B \mid R)=\frac{P(E_B \cap R)}{P(R)}=\frac{\frac{6}{40}}{\frac{3}{8}}=\frac{2}{5}$

Probability that the red marble is from box $C$ is $P(E_C \mid R)$.

$\Rightarrow P(E_C \mid R)=\frac{P(E_C \cap R)}{P(R)}=\frac{\frac{8}{40}}{\frac{3}{8}}=\frac{8}{15}$

**7.** Assume that the chances of a patient having a heart attack is $40 \%$. 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 chances 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?

## Show Answer

**Solution**

Let $A, E_1$, and $E_2$ 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.

$\therefore P(A)=0.40$

$P(E_1)=P(E_2)=\frac{1}{2}$

$P(A \mid E_1)=0.40 \times 0.70=0.28$

$P(A \mid E_2)=0.40 \times 0.75=0.30$

Probability that the patient suffering a heart attack followed a course of meditation and yoga is given by $P(E_1 \mid A)$.

$ \begin{aligned} P(E_1 \mid A) & =\frac{P(E_1) P(A \mid E_1)}{P(E_1) P(A \mid E_1)+P(E_2) P(A \mid E_2)} \\ & =\frac{\frac{1}{2} \times 0.28}{\frac{1}{2} \times 0.28+\frac{1}{2} \times 0.30} \\ & =\frac{14}{29} \end{aligned} $

**8.** 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 $\frac{1}{2}$ ).

## Show Answer

**Solution**

The total number of determinants of second order with each element being 0 or 1 is (2) $=16$

The value of determinant is positive in the following cases. $ \begin{vmatrix} 1 & 0 \\ 0 & 1\end{vmatrix} \begin{vmatrix} 1 & 1 \\ 0 & 1\end{vmatrix} \begin{vmatrix} 1 & 0 \\ 1 & 1\end{vmatrix} $

$\square$ Required probability $=\frac{3}{16}$

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

$ \begin{aligned} P(A \text{ fails }) & =0.2 \\ P(B \text{ fails alone }) & =0.15 \\ P(A \text{ and } B \text{ fail }) & =0.15 \end{aligned} $

Evaluate the following probabilities

(i) $P$ (A fails $\mid B$ has failed)

(ii) $P$ (A fails alone)

## Show Answer

**Solution**

Let the event in which $A$ fails and $B$ fails be denoted by $E_A$ and $E_B$.

$P(E_A)=0.2$

$P(E_A \square E_B)=0.15$

$P(B$ fails alone $)=P(E_B)-P(E_A \square E_B)$

$\square 0.15=P(E_B)-0.15$

$\square P(E_B)=0.3$

(i) $P(E_A \mid E_B)=\frac{P(E_A \cap E_B)}{P(E_B)}=\frac{0.15}{0.3}=0.5$

(ii) $P(A$ fails alone $)=P(E_A)-P(E_A \square E_B)$

$=0.2-0.15$

$=0.05$

**10.** 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.

## Show Answer

**Solution**

Let $E_1$ and $E_2$ respectively denote the events that a red ball is transferred from bag I to II and a black ball is transferred from bag I to II.

$P(E_1)=\frac{3}{7}$ and $P(E_2)=\frac{4}{7}$

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

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

$P(A \mid E_1)=\frac{5}{10}=\frac{1}{2}$

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

$P(A \mid E_2)=\frac{4}{10}=\frac{2}{5}$

$ \begin{aligned} \therefore P(E_2 \mid A) & =\frac{P(E_2) P(A \mid E_2)}{P(E_1) P(A \mid E_1)+P(E_2) P(A \mid E_2)} \\ & =\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{aligned} $

Choose the correct answer in each of the following:

**11.** If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B \mid A)=1$, then

(A) $A \subset B$

(B) $B \subset A$

(C) $B=\phi$

(D) $A=\phi$

## Show Answer

**Solution**

$P(A) \neq 0$ and $P(B \mid A)=1$

$P(B \mid A)=\frac{P(B \cap A)}{P(A)}$

$1=\frac{P(B \cap A)}{P(A)}$

$P(A)=P(B \cap A)$

$\Rightarrow A \subset B$

Thus, the correct answer is $A$.

**12.** If $P(A \mid B)>P(A)$, then which of the following is correct :

(A) $P(B \mid A)<P(B)$

(B) $P(A \cap B)<P(A) \cdot P(B)$

(C) $P(B \mid A)>P(B)$

(D) $P(B \mid A)=P(B)$

## Show Answer

**Solution**

$ \begin{aligned} & P(A \mid B)>P(A) \\ & \Rightarrow \frac{P(A \cap B)}{P(B)}>P(A) \\ & \Rightarrow P(A \cap B)>P(A) \cdot P(B) \\ & \Rightarrow \frac{P(A \cap B)}{P(A)}>P(B) \\ & \Rightarrow P(B \mid A)>P(B) \end{aligned} $

Thus, the correct answer is $C$.

**13.** If $A$ and $B$ are any two events such that $P(A)+P(B)-P(A$ and $B)=P(A)$, then

(A) $P(B \mid A)=1$

(B) $P(A \mid B)=1$

(C) $P(B \mid A)=0$

(D) $P(A \mid B)=0$

## Show Answer

**Solution**

$ \begin{aligned} & P(A)+P(B)-P(A \text{ and } B)=P(A) \\ & \Rightarrow P(A)+P(B)-P(A \cap B)=P(A) \\ & \Rightarrow P(B)-P(A \cap B)=0 \\ & \Rightarrow P(A \cap B)=P(B) \\ & \therefore P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B)}{P(B)}=1 \end{aligned} $

Thus, the correct answer is B.