Three fair coins with faces numbered 1 and 0 | vedclass

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(B) Let $X$ be the random variable representing the sum of the numbers on the uppermost faces. The possible values of $X$ are $0, 1, 2, 3$. Since ther

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$0.75$

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(B) Let $X$ be the random variable representing the sum of the numbers on the uppermost faces. The possible values of $X$ are $0, 1, 2, 3$. Since there are $3$ coins,the total number of outcomes is $2^3 = 8$. The probability distribution is: $P(X=0) = \frac{1}{8}$ $P(X=1) = \frac{3}{8}$ $P(X=2) = \frac{3}{8}$ $P(X=3) = \frac{1}{8}$ $E(X) = \sum x_i p_i = 0 	imes \frac{1}{8} + 1 	imes \frac{3}{8} + 2 	imes \frac{3}{8} + 3 	imes \frac{1}{8} = \frac{0+3+6+3}{8} = \frac{12}{8} = 1.5$. $E(X^2) = \sum x_i^2 p_i = 0^2 	imes \frac{1}{8} + 1^2 	imes \frac{3}{8} + 2^2 	imes \frac{3}{8} + 3^2 	imes \frac{1}{8} = \frac{0+3+12+9}{8} = \frac{24}{8} = 3$. $Variance(X) = E(X^2) - [E(X)]^2 = 3 - (1.5)^2 = 3 - 2.25 = 0.75$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$2p$	$2p$	$3p$	$p^2$	$2p^2$	$7p^2$	$2p$

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

Three fair coins with faces numbered $1$ and $0$ are tossed simultaneously. The variance $(X)$ of the probability distribution of random variable $X$,where $X$ is the sum of numbers on the uppermost faces,is

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