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

Vedclass · Accepted Answer

$\frac{9}{4}$

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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. The possible values for $X$ are $5, 2,$ and $1$.The probability distribution of $X$ is:$X$$5$$2$$1$$P(X)$$\frac{1}{4}$$\frac{1}{2}$$\frac{1}{4}$$1$. Calculate the Mean (Expected Value) $\mu$:$\mu = \sum XP(X) = (5 	imes \frac{1}{4}) + (2 	imes \frac{1}{2}) + (1 	imes \frac{1}{4}) = \frac{5}{4} + \frac{4}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2}$.$2$. Calculate $\sum X^2P(X)$:$\sum X^2P(X) = (25 	imes \frac{1}{4}) + (4 	imes \frac{1}{2}) + (1 	imes \frac{1}{4}) = \frac{25}{4} + \frac{8}{4} + \frac{1}{4} = \frac{34}{4} = \frac{17}{2}$.$3$. Calculate the Variance:$	ext{Variance}(X) = \sum X^2P(X) - [\sum XP(X)]^2 = \frac{17}{2} - (\frac{5}{2})^2 = \frac{17}{2} - \frac{25}{4} = \frac{34 - 25}{4} = \frac{9}{4}$.

$A$ 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. The variance of his winning amount is:

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