A player tosses 2 fair coins. He wins $₹ 5$ if 2 heads appear, ₹ 2 if one head appears and ₹ 1 if no head…
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
6
$\frac{5}{2}$
$\frac{9}{4}$
$\frac{17}{2}$
Solution
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 $\mathrm{P}(\mathrm{X}=1)=\frac{1}{4}$
$\therefore \quad$ 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}
$\begin{aligned}
\text { Variance of } \mathrm{X} & =\sum \mathrm{X}^2 \mathrm{P}(\mathrm{X})-[\Sigma \mathrm{XP}(\mathrm{X})]^2 \\
& =\left[\frac{25}{4}+2+\frac{1}{4}\right]-\left[\frac{5}{4}+1+\frac{1}{4}\right]^2 \\
& =\frac{34}{4}-\left(\frac{10}{4}\right)^2 \\
& =\frac{34}{4}-\frac{25}{4} \\
& =\frac{9}{4}
\end{aligned}$