In a game,two players A and B take turns thro | vedclass

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(D) Let $p_A$ be the probability of $A$ throwing a $6$,so $p_A = \frac{5}{36}$.Let $p_B$ be the probability of $B$ throwing a $7$,so $p_B = \frac{6}{3

Vedclass · Accepted Answer

$\frac{30}{61}$

Vedclass · Answer

(D) Let $p_A$ be the probability of $A$ throwing a $6$,so $p_A = \frac{5}{36}$.Let $p_B$ be the probability of $B$ throwing a $7$,so $p_B = \frac{6}{36} = \frac{1}{6}$.Let $q_A = 1 - p_A = \frac{31}{36}$ and $q_B = 1 - p_B = \frac{5}{6}$.$A$ wins if he gets a $6$ on his $1^{st}$ turn,or if both fail on their $1^{st}$ turns and $A$ gets a $6$ on his $2^{nd}$ turn,and so on.$P(A 	ext{ wins}) = p_A + (q_A q_B) p_A + (q_A q_B)^2 p_A + \dots$This is a geometric series with first term $a = p_A = \frac{5}{36}$ and common ratio $r = q_A q_B = \frac{31}{36} 	imes \frac{5}{6} = \frac{155}{216}$.$P(A 	ext{ wins}) = \frac{a}{1-r} = \frac{5/36}{1 - 155/216} = \frac{5/36}{61/216} = \frac{5}{36} 	imes \frac{216}{61} = \frac{5 	imes 6}{61} = \frac{30}{61}$.

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