Question

A dice is thrown twice. A success is an even number on each throw. Find the probability distribution of the number of successes.

Answer 1

Total number of cases on throwing a dice
$S=\{1,2,3,4,5,6\}$
$n\left(S\right)=\mathrm{6......}\left(1\right)$
number of favourable cases
$E=\{2,4,6,\}$
$n\left(E\right)=\mathrm{3....}\left(2\right)$
$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{6}=\frac{1}{2}$
$P=\frac{1}{2}$
probability of failure $=9=1-\frac{1}{2}=\frac{1}{2}$
$n=2$
for $x=0$ (no success)
$P(x=0)={{}^{2}C}_0{\left(\frac{1}{2}\right)}^0{\left(\frac{1}{2}\right)}^{2-0}={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
for $x=1$ (one success)
$P(x=1)={{}^{2}C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^{2-1}=\frac{1}{2}$
for $x=2$ (two successes)
$P(x=2)={{}^{2}C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{2-2}=\frac{1}{4}$
Probability distribution

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

$\sum P(x=r)=\frac{1}{4}+\frac{1}{2}+\frac{1}{4}=1$