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) $A \cap B=\phi$

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

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

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} $

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 $=\frac{5}{5.25}=\frac{20}{21}$

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)

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

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 $=\frac{2}{7}$

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 $P(E_C \mid R)$.

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

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} $

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 $=\frac{3}{16}$

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

$=0.2-0.15$

$=0.05$

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:

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 $A$.

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 $C$.

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.