Suppose three coins are tossed simultaneously | vedclass

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(B) Let $X$ denote the number of heads. When three coins are tossed,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$,so the total

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$X=x$$0$$1$$2$$3$$P(X=x)$$\frac{1}{8}$$\frac{3}{8}$$\frac{3}{8}$$\frac{1}{8}$

Vedclass · Answer

(B) Let $X$ denote the number of heads. When three coins are tossed,the sample space is $S = \{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT\}$,so the total number of outcomes is $8$.The possible values of $X$ are $0, 1, 2,$ and $3$.$P(X=0) = P(\{TTT\}) = \frac{1}{8}$$P(X=1) = P(\{HTT, THT, TTH\}) = \frac{3}{8}$$P(X=2) = P(\{HHT, HTH, THH\}) = \frac{3}{8}$$P(X=3) = P(\{HHH\}) = \frac{1}{8}$Thus,the probability distribution is:$X=x$$0$$1$$2$$3$$P(X=x)$$\frac{1}{8}$$\frac{3}{8}$$\frac{3}{8}$$\frac{1}{8}$Therefore,Option $(B)$ is the correct answer.

Suppose three coins are tossed simultaneously. If $X$ denotes the number of heads,then the probability distribution of $X$ is:

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$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{2}{8}$	$\frac{2}{8}$	$\frac{3}{8}$	$\frac{1}{8}$