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(D) Let $H$ denote Head and $T$ denote Tail. The experiment stops when $H$ appears or $T$ appears $4$ times in succession.For $X=1$: The outcome is ${

Vedclass · Accepted Answer

$X$ $1$ $2$ $3$ $4$     $P(X=x)$ $\frac{1}{2}$ $\frac{1}{4}$ $\frac{1}{8}$ $\frac{1}{8}$

Vedclass · Answer

(D) Let $H$ denote Head and $T$ denote Tail. The experiment stops when $H$ appears or $T$ appears $4$ times in succession.For $X=1$: The outcome is ${H}$. $P(X=1) = \frac{1}{2}$.For $X=2$: The outcome is ${TH}$. $P(X=2) = \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{4}$.For $X=3$: The outcome is ${TTH}$. $P(X=3) = \frac{1}{2} 	imes \frac{1}{2} 	imes \frac{1}{2} = \frac{1}{8}$.For $X=4$: The outcomes are ${TTTH, TTTT}$. $P(X=4) = (\frac{1}{2})^4 + (\frac{1}{2})^4 = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}$.Thus,the distribution is:$P(X=1) = \frac{1}{2}$,$P(X=2) = \frac{1}{4}$,$P(X=3) = \frac{1}{8}$,$P(X=4) = \frac{1}{8}$.This matches option $D$.

$A$ coin is tossed until one head appears or a tail appears $4$ times in succession. The probability distribution of the number of tosses $X$ is:

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