Question

A die is tossed twice. A 'success' is getting an odd number on a toss. Find the variance of the number of successes.

Answer 1

It is given that "success" denotes the event of getting the numbers 1, 3 or 5. Then,
$P\left(\mathrm{success}\right)=\frac{1}{2}$

Also, "failure" denotes the event of getting the numbers 2, 4 or 6. Then,
$P\left(\mathrm{failure}\right)=\frac{1}{2}$

Let X denote the event of getting success.Then, X can take the values 0, 1 and 2.
Now,
$$\begin{gathered} P\left(X=0\right)=P\left(\mathrm{no}\mathrm{success}\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4} \\ P\left(X=1\right)=P\left(1\mathrm{success}\right)=\left(\frac{1}{2}\times \frac{1}{2}\right)+\left(\frac{1}{2}\times \frac{1}{2}\right)=\frac{1}{2} \\ P\left(X=2\right)=P\left(2\mathrm{success}\right)=\frac{1}{2}\times \frac{1}{2}=\frac{1}{4} \end{gathered}$$

Thus, the probability distribution of X is given by

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

Computation of mean and variance

xi	pi	pixi	pixi2
0	$\frac{1}{4}$	0	0
1	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$\frac{1}{4}$	$\frac{1}{2}$	1
		${\sum }_{}^{}$pixi = 1	${\sum }_{}^{}$pixi2 = $\frac{3}{2}$

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=1 \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2=\frac{3}{2}-1=\frac{1}{2} \end{gathered}$$