A player tosses 2 fair coins. He wins ₹ \(5\) if 2 heads appear, ₹ 2 if one head appears and ₹ 1 if no head appears. Then the variance of his winning amount in ₹ is

When player tosses 2 fair coins, then \(\mathrm{S}=\{\mathrm{HH}, \mathrm{HT}, \mathrm{TH}, \mathrm{TT}\}\)
Let \(\mathrm{X}\) be a random variable that denotes the amount received by player.
Then, \(X\) can take values 5,2 and 1
Now, \(\mathrm{P}(\mathrm{X}=5)=\frac{1}{4}, \mathrm{P}(\mathrm{X}=2)=\frac{1}{2}\) and \(\mathrm{P}(\mathrm{X}=1)=\frac{1}{4}\)
\(\therefore\) The probability distribution of \(\mathrm{X}\) is as follows:
\(\begin{array}{|c|c|c|c|}
\hline \mathrm{X} & 5 & 2 & 1 \\
\hline \mathrm{P}(\mathrm{X}) & \frac{1}{4} & \frac{1}{2} & \frac{1}{4} \\
\hline
\end{array}\)
\(\text { Variance of } \mathrm{X} =\sum \mathrm{X}^2 \mathrm{P}(\mathrm{X})-[\Sigma \mathrm{XP}(\mathrm{X})]^2 \)
\( =\left[\frac{25}{4}+2+\frac{1}{4}\right]-\left[\frac{5}{4}+1+\frac{1}{4}\right]^2 \)
\( =\frac{34}{4}-\left(\frac{10}{4}\right)^2 \)
\( =\frac{34}{4}-\frac{25}{4} \)
\( =\frac{9}{4}\)