If two natural numbers $r$ and $s$ are drawn one at a time, without replacement from the set $S=\{1,2,3, \ldots n\}$, then find $P(r \leq p / s \leq p)$, where $p \in S$.
$$\begin{aligned} & \because \text { Set } S=\{1,2,3, \ldots, n\} \\ & \therefore \quad P(r \leq p / S \leq p)=\frac{P(p \cap S)}{P(S)} \\ & \\ \end{aligned}$$
$=\frac{p-1}{n} \times \frac{n}{n-1}=\frac{p-1}{n-1}$
Find the probability distribution of the maximum of the two scores obtained when a die is thrown twice. Determine also the mean of the distribution.
Let $X$ is the random variable score obtained when a die is thrown twice.
$$\begin{array}{lrl} \therefore & X & =1,2,3,4,5,6 \\ \text { Here, } & S & =\{(1,1),(1,2),(2,1),(2,2),(1,3),(2,3),(3,1),(3,2),(3,3), \ldots,(6,6)\} \\ \therefore & P(X & =1)=\frac{1}{6} \cdot \frac{1}{6}=\frac{1}{36} \\ & & P(X=2)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{3}{36} \\ & P(X & =3)=\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}+\frac{1}{6} \cdot \frac{1}{6}=\frac{5}{36} \end{array}$$
$$\begin{aligned} \text{Similarly,}\quad & P(X=4)=\frac{7}{36} \\ & P(X=5)=\frac{9}{36} \\ & P(X=6)=\frac{11}{36} \end{aligned}$$
So, the required distribution is,
| $X$ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| $P(X)$ | 1/36 | 3/36 | 5/36 | 7/36 | 9/36 | 11/36 |
$$\begin{aligned} &\text { Also, we know that, Mean }\{E(X)\}=\Sigma X P(X)\\ &=\frac{1}{36}+\frac{6}{36}+\frac{15}{36}+\frac{28}{36}+\frac{45}{36}+\frac{66}{36}=\frac{161}{36} \end{aligned}$$
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 $=P(\bar{A}) \cdot P(\bar{B}) \cdot P(A)=\frac{31}{36} \cdot \frac{5}{6} \cdot \frac{5}{36}=\frac{775}{216 \cdot 36}=\frac{775}{7776}$