A die is tossed twice. If a 'success' is getting an even number on a toss, then find the variance of the number of successes.
Let $X$ be the random variable for a 'success' for getting an even number on a toss.
$\therefore \quad X=0,1,2, n=2, p=\frac{3}{6}=\frac{1}{2}$ and $q=\frac{1}{2}$
$$\begin{array}{ll} \text { At } X=0, & P(X=0)={ }^2 C_0\left(\frac{1}{2}\right)^0\left(\frac{1}{2}\right)^{2-0}=\frac{1}{4} \\ \text { At } X=1, & P(X=1)={ }^2 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^{2-1}=2 \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{2} \\ \text { At } X=2, & P(X=2)={ }^2 C_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^{2-2}=\frac{1}{4} \end{array}$$
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $\frac{1}{4}$ | $\frac{1}{2}$ | $\frac{1}{4}$ |
| $XP(X)$ | 0 | $\frac{1}{2}$ | $\frac{1}{2}$ |
| $X^2P(X)$ | 0 | $\frac{1}{2}$ | 1 |
$\therefore \quad \Sigma X P(X)=0+\frac{1}{2}+\frac{1}{2}=1\quad\text{.... (i)}$
and $$\Sigma X^2 P(X)=0+\frac{1}{2}+1=\frac{3}{2}\quad\text{.... (ii)}$$
$$\begin{aligned} \because \operatorname{Var}(X) & =E\left(X^2\right)-[E(X)]^2 \\ & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2=\frac{3}{2}-(1)^2=\frac{1}{2} \quad \text { [using Eqs. (i) and (ii)] } \end{aligned}$$
There are 5 cards numbered 1 to 5, one number on one card. Two cards are drawn at random without replacement. Let $X$ denotes the sum of the numbers on two cards drawn. Find the mean and variance of $X$.
Here, $$S=\{(1,2),(2,1),(1,3),(3,1),(2,3),(3,2),(1,4),(4,1),(1,5),(5,1),(2,4),(4,2),(2,5),(5,2),(3,4),(4,3),(3,5),(5,3),(5,4),(4,5)\}$$
$$\Rightarrow n(S)=20$$
Let random variable be $X$ which denotes the sum of the numbers on two cards drawn.
$\therefore\qquad X=3,4,5,6,7,8,9$
$$\begin{aligned} & \text { At } X=3, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=4, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=5, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=6, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=7, P(X)=\frac{4}{20}=\frac{1}{5} \\ & \text { At } X=8, P(X)=\frac{2}{20}=\frac{1}{10} \\ & \text { At } X=9, P(X)=\frac{2}{20}=\frac{1}{10} \end{aligned}$$
$\therefore \quad$ Mean, $E(X)=\Sigma X P(X)=\frac{3}{10}+\frac{4}{10}+\frac{5}{5}+\frac{6}{5}+\frac{7}{5}+\frac{8}{10}+\frac{9}{10}$
$=\frac{3+4+10+12+14+8+9}{10}=6$
$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} \Sigma X^2 P(X) & =\frac{9}{10}+\frac{16}{10}+\frac{25}{5}+\frac{36}{5}+\frac{49}{5}+\frac{64}{10}+\frac{81}{10} \\ & =\frac{9+16+50+72+98+64+81}{10}=39 \end{aligned} \end{aligned}$$
$$\begin{aligned} \therefore \quad \operatorname{Var}(X) & =\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =39-(6)^2=39-36=3 \end{aligned}$$
If $P(A)=\frac{4}{5}$ and $P(A \cap B)=\frac{7}{10}$, then $P(B / A)$ is equal to
If $P(A \cap B)=\frac{7}{10}$ and $P(B)=\frac{17}{20}$, then $P(A / B)$ equals to
If $P(A)=\frac{3}{10}, P(B)=\frac{2}{5}$ and $P(A \cup B)=\frac{3}{5}$, then $P(B / A)+P(A / B)$ equals to