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 $S=\left\{HT,TH,T,T,HH\right\}$
Let $X$ be a random variable that denotes the amount received by the player. Then, $x$ can take value $5,2$ and 1.
Now, $P(X=5)=\frac{1}{4},P(X=2)=\frac{2}{4}=\frac{1}{2}$
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 $X=\Sigma X^{2}P(X)-[\Sigma XP(X){]}^2$
Now, $\Sigma XP(X)=5\times \frac{1}{4}+2\times \frac{1}{2}+1\times \frac{1}{4}$
$=\frac{5}{4}+\frac{4}{4}+\frac{1}{4}=\frac{10}{4}=\frac{5}{2}$
$\therefore$ Variance of $X=\Sigma X^{2}P(X)-{\left(\Sigma XP(X)\right)}^2$
$=\left(25\times \frac{1}{4}+4\times \frac{1}{2}+1\times \frac{1}{4}\right)-{\left(\frac{5}{2}\right)}^2$
$=\left(\frac{25+8+1}{4}\right)-\frac{25}{4}=\frac{9}{4}$