Question

A die is tossed twice. A success is getting an odd number on a toss. Find the mean and the variance of the probability distribution of the number of successes.

Answer 1

Let $x$ be the random variable denoting the number of times an odd number (the number of successes) when a die is tossed twice.

Then $x$ takes the values $0,1,2$

Let $P(X=0)$ be probability of getting no odd number (both times showing even).

$\therefore P(X=0)=\frac{3}{6}\times \frac{3}{6}=\frac{1}{4}$

Let $P(X=1)$ be probability of getting odd number once.

$\therefore P(X=1)=2C_1\frac{3}{6}\times \frac{3}{6}=\frac{6}{6}\times \frac{3}{6}=\frac{1}{2}$

Let $P(X=2)$ be probability of getting odd number twice.

$\therefore P(X=2)=\frac{3}{6}\times \frac{3}{6}=\frac{1}{4}$

Thus the probability distribution of $X$ is given by

$X=x$	$x=0$	$x=1$	$x=2$
$P(X=x)$	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

We know that mean $E\left(X\right)=\sum x_i{P}_i=0\times \frac{1}{4}+1\times \frac{1}{2}+2\times \frac{1}{4}$

$\therefore E\left(X\right)=0+\frac{1}{2}+\frac{1}{2}=1$

Thus mean $E\left(X\right)=1$.

We know that $var\left(X\right)=E\left(X^2\right)-\left[E\right(X){]}^2$

$E\left(X^2\right)=\sum x_i^2{P}_i=0\times \frac{1}{4}+1^2\times \frac{1}{2}+2^2\times \frac{1}{4}$

$\therefore E\left(X^2\right)=0+\frac{1}{2}+4\times \frac{1}{4}=\frac{3}{2}$

Thus $var\left(X\right)=\frac{3}{2}-[1{]}^2=\frac{3}{2}-1=\frac{1}{2}$

Hence mean is $1$ and variance is $\frac{1}{2}$.