A player tosses 2 fair coins. He wins Rs. 5 i | vedclass

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(A) The sample space for tossing $2$ coins is $\{HH, HT, TH, TT\}$.The probabilities are:$P(X=5) = P(HH) = \frac{1}{4}$$P(X=2) = P(HT, TH) = \frac{2}{

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$\frac{9}{4}$

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(A) The sample space for tossing $2$ coins is $\{HH, HT, TH, TT\}$.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}$We calculate the mean $E(X) = \sum p_i x_i = (5 	imes \frac{1}{4}) + (2 	imes \frac{1}{2}) + (1 	imes \frac{1}{4}) = \frac{5}{4} + 1 + \frac{1}{4} = \frac{6}{4} + 1 = \frac{3}{2} + 1 = \frac{5}{2}$.Next,we calculate $E(X^2) = \sum p_i x_i^2 = (5^2 	imes \frac{1}{4}) + (2^2 	imes \frac{1}{2}) + (1^2 	imes \frac{1}{4}) = \frac{25}{4} + 2 + \frac{1}{4} = \frac{26}{4} + 2 = \frac{13}{2} + 2 = \frac{17}{2}$.Variance $Var(X) = E(X^2) - [E(X)]^2 = \frac{17}{2} - (\frac{5}{2})^2 = \frac{17}{2} - \frac{25}{4} = \frac{34-25}{4} = \frac{9}{4}$.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

$X = x$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x)$	$0.2$	$0.3$	$0.1$	$0.15$	$0.25$

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

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