If three fair coins are tossed,then the varia | vedclass

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(C) Let the random variable $X$ denote the number of heads obtained when three fair coins are tossed. The probability distribution of $X$ is given by

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$0.75$

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(C) Let the random variable $X$ denote the number of heads obtained when three fair coins are tossed. The probability distribution of $X$ is given by the following table:| $X = x$ | $0$ | $1$ | $2$ | $3$ || :--- | :--- | :--- | :--- | :--- || $P(X = x)$ | $\frac{1}{8}$ | $\frac{3}{8}$ | $\frac{3}{8}$ | $\frac{1}{8}$ |The expected value $E(X)$ is calculated as:$E(X) = \sum x P(X=x) = 0 	imes \frac{1}{8} + 1 	imes \frac{3}{8} + 2 	imes \frac{3}{8} + 3 	imes \frac{1}{8} = 0 + \frac{3}{8} + \frac{6}{8} + \frac{3}{8} = \frac{12}{8} = \frac{3}{2} = 1.5$The expected value $E(X^2)$ is calculated as:$E(X^2) = \sum x^2 P(X=x) = 0^2 	imes \frac{1}{8} + 1^2 	imes \frac{3}{8} + 2^2 	imes \frac{3}{8} + 3^2 	imes \frac{1}{8} = 0 + \frac{3}{8} + \frac{12}{8} + \frac{9}{8} = \frac{24}{8} = 3$The variance $V(X)$ is given by the formula:$V(X) = E(X^2) - [E(X)]^2$$V(X) = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12 - 9}{4} = \frac{3}{4} = 0.75$

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0$	$2k$	$k$	$3k$	$2k^2$	$2k$	$k^2+k$	$7k^2$

If three fair coins are tossed,then the variance of the number of heads obtained is

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