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(B) Three fair coins numbered $1$ and $0$ are tossed. The sample space $S = \{111, 110, 101, 011, 100, 010, 001, 000\}$. The total number of outcomes

Vedclass · Accepted Answer

$0.75$

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(B) Three fair coins numbered $1$ and $0$ are tossed. The sample space $S = \{111, 110, 101, 011, 100, 010, 001, 000\}$. The total number of outcomes $n(S) = 8$. $X$ represents the sum of the numbers on the uppermost faces. The probability distribution of $X$ is: $X$$0$$1$$2$$3$$P(X)$$1/8$$3/8$$3/8$$1/8$ $E(X) = \sum x_i P(x_i) = 0 	imes \frac{1}{8} + 1 	imes \frac{3}{8} + 2 	imes \frac{3}{8} + 3 	imes \frac{1}{8} = \frac{0+3+6+3}{8} = \frac{12}{8} = \frac{3}{2}$. $E(X^2) = \sum x_i^2 P(x_i) = 0^2 	imes \frac{1}{8} + 1^2 	imes \frac{3}{8} + 2^2 	imes \frac{3}{8} + 3^2 	imes \frac{1}{8} = \frac{0+3+12+9}{8} = \frac{24}{8} = 3$. $\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = 3 - (\frac{3}{2})^2 = 3 - \frac{9}{4} = \frac{12-9}{4} = \frac{3}{4} = 0.75$.

Three fair coins,each numbered $1$ and $0$,are tossed simultaneously. The variance $\operatorname{Var}(X)$ of the probability distribution of the random variable $X$,where $X$ is the sum of the numbers on the uppermost faces,is:

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