The random variable $X$ can take only the values $0,1,2$. If
$$P(X=0)=P(X=1)=p \text { and } E\left(X^2\right)=E[X]$$
then find the value of $p$.
Since, $X=0,1,2$ and $P(X)$ at $X=0$ and 1 is $p$, let at $X=2, P(X)$ is $x$.
$$\begin{aligned} \Rightarrow & \quad p+p+x =1 \\ \Rightarrow & \quad x =1-2 p \end{aligned}$$
We get, the following distribution.
| $X$ | 0 | 1 | 2 |
|---|---|---|---|
| $P(X)$ | $p$ | $p$ | $1-2p$ |
$$\begin{aligned} \therefore\quad E[X] & =\Sigma X P(X) \\ & =0 \cdot p+1 \cdot p+2(1-2 p) \\ & =p+2-4 p=2-3 p \end{aligned}$$
$$\begin{aligned} &\text { and }\\ &\begin{aligned} & E\left(X^2\right)=\Sigma X^2 P(X) \\ & =0 \cdot p+1 \cdot p+4 \cdot(1-2 p) \\ & =p+4-8 p=4-7 p \end{aligned} \end{aligned}$$
$$\begin{array}{lrl} \text { Also, } & \text { given that } E\left(X^2\right) & =E[X] \\ \Rightarrow & 4-7 p & =2-3 p \\ \Rightarrow & 4 p & =2 \Rightarrow p=\frac{1}{2} \end{array}$$
Find the variance of the following distribution.
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $P(X)$ | $\frac{1}{6}$ | $\frac{5}{18}$ | $\frac{2}{9}$ | $\frac{1}{6}$ | $\frac{1}{9}$ | $\frac{1}{18}$ |
We have,
| $X$ | 0 | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|---|
| $P(X)$ | $\frac{1}{6}$ | $\frac{5}{18}$ | $\frac{2}{9}$ | $\frac{1}{6}$ | $\frac{1}{9}$ | $\frac{1}{18}$ |
| $XP(X)$ | 0 | $\frac{5}{18}$ | $\frac{4}{9}$ | $\frac{1}{2}$ | $\frac{4}{9}$ | $\frac{5}{18}$ |
| $X^2P(X)$ | 0 | $\frac{5}{18}$ | $\frac{8}{9}$ | $\frac{3}{2}$ | $\frac{16}{9}$ | $\frac{25}{18}$ |
$$\begin{aligned} \therefore \quad \text { Variance } & =E\left(X^2\right)-[E(X)]^2=\Sigma X^2 P(X)-[\Sigma X P(X)]^2 \\ & =\left[0+\frac{5}{18}+\frac{8}{9}+\frac{3}{2}+\frac{16}{9}+\frac{25}{18}\right]-\left[0+\frac{5}{18}+\frac{4}{9}+\frac{1}{2}+\frac{4}{9}+\frac{5}{18}\right]^2 \\ & =\left[\frac{5+16+27+32+25}{18}\right]-\left[\frac{5+8+9+8+5}{18}\right]^2 \\ & =\frac{105}{18}-\frac{35 \cdot 35}{18 \cdot 18}=\frac{18 \cdot 105-35 \cdot 35}{18 \cdot 18} \\ & =\frac{35}{18 \cdot 18}[54-35]=\frac{19 \cdot 35}{324}=\frac{665}{324} \end{aligned}$$
$A$ and $B$ throw a pair of dice alternately. A wins the game, if he gets a total of 6 and $B$ wins, if she gets a total of 7 . If $A$ starts the game, then find the probability of winning the game by $A$ in third throw of the pair of dice.
Let $A_1=A$ total of $6=\{(2,4),(1,5),(5,1),(4,2),(3,3)\}$
and $B_1=$ A total of $7=\{(2,5),(1,6),(6,1),(5,2),(3,4),(4,3)\}$
Let $P(A)$ is the probability, if $A$ wins in a throw $\Rightarrow P(A)=\frac{5}{36}$
and $P(B)$ is the probability, if $B$ wins in a throw $\Rightarrow P(B)=\frac{1}{6}$
$\therefore$ Required probability $=P(\bar{A}) \cdot P(\bar{B}) \cdot P(A)=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{216 \cdot 36}=\frac{775}{7776}$
Two dice are tossed. Find whether the following two events $A$ and $B$ are independent $A=\{(x, y): x+y=11\}$ and $B=\{(x, y): x \neq 5\}$, where $(x, y)$ denotes a typical sample point.
$$\begin{aligned} & \text { We have, } A=\{(x, y): x+y=11\} \text { and } B=\{(x, y): x \neq 5\} \\ & \begin{aligned} \therefore & A=\{(5,6),(6,5)\}, B=\{(1,1),(1,2),(1,3),(1,4),(1,5)(1,6),(2,1),(2,2),(2,3),(2,4) \\ & (2,5)(2,6),(3,1),(3,2),(3,3),(3,4),(3,5)(3,6),(4,1),(4,2),(4,3),(4,4),(4,5)(4,6), \\ & (6,1),(6,2),(6,3),(6,4),(6,5),(6,6)\} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\begin{array}{ll} \Rightarrow & n(A)=2, n(B)=30 \text { and } n(A \cap B)=1 \\ \therefore & P(A)=\frac{2}{36}=\frac{1}{18} \text { and } P(B)=\frac{30}{36}=\frac{5}{6} \\ \Rightarrow & P(A) \cdot P(B)=\frac{5}{108} \text { and } P(A \cap B)=\frac{1}{36} \neq P(A) \cdot P(B) \end{array}\\ &\text { So, } A \text { and } B \text { are not independent. } \end{aligned}$$
Q. 40 An urn contains $m$ white and $n$ black balls. A ball is drawn at random and is put back into the urn along with $k$ additional balls of the same colour as that of the ball drawn. A ball is again drawn at random. Show that the probability of drawing a white ball now does not depend on $k$.
$$\begin{aligned} \text{Let }\quad U & =\{m \text { white, } n \text { black balls }\} \\ E_1 & =\{\text { First ball drawn of white colour }\} \\ E_2 & =\{\text { First ball drawn of black colour }\} \end{aligned}$$
$$\begin{aligned} &\text { and } \quad E_3=\{\text { Second ball drawn of white colour }\}\\ &\therefore \quad P\left(E_1\right)=\frac{m}{m+n} \text { and } P\left(E_2\right)=\frac{n}{m+n} \end{aligned}$$
$$\begin{aligned} &\text { Also, }\\ &\begin{aligned} P\left(E_3 / E_1\right) & =\frac{m+k}{m+n+k} \text { and } P\left(E_3 / E_2\right)=\frac{m}{m+n+k} \\ \therefore\quad P\left(E_3\right) & =P\left(E_1\right) \cdot P\left(E_3 / E_1\right)+P\left(E_2\right) \cdot P\left(E_3 / E_2\right) \\ & =\frac{m}{m+n} \cdot \frac{m+k}{m+n+k}+\frac{n}{m+n} \cdot \frac{m}{m+n+k} \\ & =\frac{m(m+k)+n m}{(m+n+k)(m+n)}=\frac{m^2+m k+n m}{(m+n+k)(m+n)} \\ & =\frac{m(m+k+n)}{(m+n+k)(m+n)}=\frac{m}{m+n} \end{aligned} \end{aligned}$$
Hence, the probability of drawing a white ball does not depend on k.