When player tosses 2 fair coins then
$S=\left\{HT,TH,TT,HH\right\}$
Let X be a random variable that denotes the amount received by the player.
Then, X can take value 5, 2 and 1.
Now, $P\left(X=5\right)=\frac{1}{4},P\left(X=2\right)=\frac{2}{4}=\frac{1}{2}$
and $P\left(X=1\right)=\frac{1}{4}$
Thus, the probability distribution of X is
$\stackrel{-}{\begin{array}{c}\underset{\overline{}}{X521}\\ P\left(X\right)\frac{1}{4}\frac{1}{2}\frac{1}{4}\end{array}}$
$\therefore$ Variance of $X={\sum X^{2}P\left(X\right)-\left[\sum XP\left(X\right)\right]}^2$
Now $\sum XP\left(X\right)=5\times \frac{1}{4}+2\times \frac{1}{2}+1\times \frac{1}{4}$
$=\frac{5}{4}+\frac{4}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$
$\therefore$ Variance of $X=\sum X^{2}P\left(X\right)-{\left(\sum XP\left(X\right)\right)}^2$
$=\left(25\times \frac{1}{4}+4\times \frac{1}{2}+1\times \frac{1}{4}\right)-{\left(\frac{5}{2}\right)}^2$
$=\left(\frac{25+8+1}{4}\right)-\frac{25}{4}=\frac{9}{4}$