Question

A player tosses $2$ fair coins. He wins Rs. $5$ if $2$ heads appear, Rs. $2$ If $1$ head appear and Rs.$1$ if no head appears, then variance of his winning amount is

• A. $6$
• B. $\frac{5}{2}$
• C. $\frac{9}{4}$
• D. $\frac{17}{2}$

Answer 1

Answer: (C)

When player tosses 2 fair coins then $S=\left\{H T, T H, T,T, H H\right\}$
Let $X$ be a random variable that denotes the amount received by the player. Then, $x$ can take value $5, 2$ and 1.
Now, $P(X=5)=\frac{1}{4}, P(X=2)=\frac{2}{4}=\frac{1}{2}$
and $P(X=1)=\frac{1}{4}$
Thus, the probability distribution of $X$ is

X	5	2	1
$P(X)$	$\frac{1}{4}$	$\frac{1}{2}$	$\frac{1}{4}$

$\therefore $ Variance of $X=\Sigma X^{2} P(X)-[\Sigma X P(X)]^{2}$
Now, $\Sigma X P(X) =5 \times \frac{1}{4}+2 \times \frac{1}{2}+1 \times \frac{1}{4}$
$=\frac{5}{4}+\frac{4}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$
$\therefore $ Variance of $X=\Sigma X^{2} P(X)-\left(\Sigma X P(X)\right)^{2}$
$=\left(25 \times \frac{1}{4}+4 \times \frac{1}{2}+1 \times \frac{1}{4}\right)-\left(\frac{5}{2}\right)^{2}$
$=\left(\frac{25+8+1}{4}\right)-\frac{25}{4}=\frac{9}{4}$