Two players A and B alternatively toss 3 coin | vedclass

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(C) Let $p$ be the probability of getting $2$ heads and $1$ tail in a single toss of $3$ coins. The total outcomes are $2^3 = 8$. The favorable outcom

Vedclass · Accepted Answer

$\frac{15}{39}$

Vedclass · Answer

(C) Let $p$ be the probability of getting $2$ heads and $1$ tail in a single toss of $3$ coins. The total outcomes are $2^3 = 8$. The favorable outcomes are ${HHT, HTH, THH}$, so $p = \frac{3}{8}$.The probability of not getting $2$ heads and $1$ tail is $q = 1 - p = 1 - \frac{3}{8} = \frac{5}{8}$.Player $A$ starts. $B$ wins if $A$ fails, then $B$ succeeds, or if $A$ fails, $B$ fails, $A$ fails, then $B$ succeeds, and so on.The probability that $B$ wins is $P(B) = qp + q^3p + q^5p + \dots$This is a geometric series with first term $a = qp = \frac{5}{8} 	imes \frac{3}{8} = \frac{15}{64}$ and common ratio $r = q^2 = (\frac{5}{8})^2 = \frac{25}{64}$.The sum of the infinite geometric series is $S = \frac{a}{1 - r} = \frac{15/64}{1 - 25/64} = \frac{15/64}{39/64} = \frac{15}{39}$.

Two players $A$ and $B$ alternatively toss $3$ coins simultaneously. The player who gets $2$ heads and $1$ tail first, wins the game. If the game continues until someone wins and if $A$ begins the game, the probability that $B$ wins the game is

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