A player tosses 2 fair coins. He wins ₹ 5 if | vedclass

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(C) When a player tosses $2$ fair coins,the sample space is $S = \{HH, HT, TH, TT\}$.Let $X$ be a random variable that denotes the amount received by

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$\frac{9}{4}$

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(C) When a player tosses $2$ fair coins,the sample space is $S = \{HH, HT, TH, TT\}$.Let $X$ be a random variable that denotes the amount received by the player.The possible values for $X$ are $5, 2,$ and $1$.The probabilities are:$P(X=5) = P(\{HH\}) = \frac{1}{4}$$P(X=2) = P(\{HT, TH\}) = \frac{2}{4} = \frac{1}{2}$$P(X=1) = P(\{TT\}) = \frac{1}{4}$The probability distribution is:$E(X) = \sum X P(X) = 5 	imes \frac{1}{4} + 2 	imes \frac{1}{2} + 1 	imes \frac{1}{4} = \frac{5}{4} + 1 + \frac{1}{4} = \frac{10}{4} = 2.5$$E(X^2) = \sum X^2 P(X) = 5^2 	imes \frac{1}{4} + 2^2 	imes \frac{1}{2} + 1^2 	imes \frac{1}{4} = \frac{25}{4} + 2 + \frac{1}{4} = \frac{34}{4} = 8.5$$	ext{Variance}(X) = E(X^2) - [E(X)]^2 = \frac{34}{4} - \left(\frac{10}{4}ight)^2 = \frac{34}{4} - \frac{100}{16} = \frac{136 - 100}{16} = \frac{36}{16} = \frac{9}{4}$.

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

$A$ 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

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