Question

Find the mean and variance of the number of tails when three coins are tossed simultaneously.

Answer 1

$A\text{coin is tossed 3 times, hence total number of cases =8}$
$\text{S:}\left\{\text{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT}\right\}$
$\text{i) when no tail (X=0)}$
$\text{P(X)=1/8}$
$\text{ii) When 1 tail (X=1)}$
$\text{P}\left(X\right)=3/8$
$iii)when2tails(X=2)$
$P\left(X\right)=3/8$
$iv)when3tails(X=3)$
$P\left(X\right)=1/8$
$Mean\left(\right)=\sum x\left(p\left(x\right)\right)=0\times \frac{1}{8}+1\times \frac{3}{8}+2\times \frac{3}{8}+3\times \frac{1}{8}$
$=\frac{12}{8}=\frac{3}{2}$
$Variance\left({\sigma }^2\right)={\sum x^{2}p\left(x\right)-[\sum xp\left(x\right)]}^2$
$\sum x^{2}p\left(x\right)=0\times \frac{1}{8}+1^2\times \frac{3}{8}+2^2\times \frac{3}{8}+3^2\times \frac{1}{8}$
$=\frac{3+12+9}{8}=3$
$plugging\mathrm{in}thevalues$
$Variance\left({\sigma }^2\right)=3-{\left(\frac{3}{2}\right)}^2$
$=3-\frac{9}{4}=\frac{3}{4}$