Two players A and B are alternately throwing | vedclass

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(B) Let $p$ be the probability of getting a head and a $6$ in a single throw. The probability of getting a head is $\frac{1}{2}$ and the probability o

Vedclass · Accepted Answer

$\frac{11}{23}$

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(B) Let $p$ be the probability of getting a head and a $6$ in a single throw. The probability of getting a head is $\frac{1}{2}$ and the probability of getting a $6$ is $\frac{1}{6}$. Since these are independent events, $p = \frac{1}{2} 	imes \frac{1}{6} = \frac{1}{12}$. The probability of not winning in a single throw is $q = 1 - p = \frac{11}{12}$.$A$ starts the game. $B$ wins if $A$ fails, then $B$ succeeds, or $A$ fails, $B$ fails, $A$ fails, $B$ succeeds, and so on.The probability that $B$ wins is:$P(B) = qp + q^3p + q^5p + \dots$This is an infinite geometric series with first term $a = qp = \frac{11}{12} 	imes \frac{1}{12} = \frac{11}{144}$ and common ratio $r = q^2 = (\frac{11}{12})^2 = \frac{121}{144}$.$P(B) = \frac{a}{1-r} = \frac{\frac{11}{144}}{1 - \frac{121}{144}} = \frac{\frac{11}{144}}{\frac{23}{144}} = \frac{11}{23}$.

Two players $A$ and $B$ are alternately throwing a coin and a die together. $A$ player who first throws a head and a $6$ wins the game. If $A$ starts the game, then the probability that $B$ wins the game is:

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