A and B play 12 games. A wins 6 times,B wins | vedclass

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(B) The probability that $A$ wins a game is $P(A) = \frac{6}{12} = \frac{1}{2}$.The probability that $B$ wins a game is $P(B) = \frac{4}{12} = \frac{1

Vedclass · Accepted Answer

$5/36$

Vedclass · Answer

(B) The probability that $A$ wins a game is $P(A) = \frac{6}{12} = \frac{1}{2}$.The probability that $B$ wins a game is $P(B) = \frac{4}{12} = \frac{1}{3}$.The probability that they win alternately in a series of $3$ games is the sum of the probabilities of the sequences $(A, B, A)$ and $(B, A, B)$.$P(	ext{alternating}) = P(A) 	imes P(B) 	imes P(A) + P(B) 	imes P(A) 	imes P(B)$$P(	ext{alternating}) = \left(\frac{1}{2} 	imes \frac{1}{3} 	imes \frac{1}{2}ight) + \left(\frac{1}{3} 	imes \frac{1}{2} 	imes \frac{1}{3}ight)$$P(	ext{alternating}) = \frac{1}{12} + \frac{1}{18} = \frac{3 + 2}{36} = \frac{5}{36}$.

$A$ and $B$ play $12$ games. $A$ wins $6$ times,$B$ wins $4$ times,and $2$ games are drawn. If $A$ and $B$ play a series of $3$ games,what is the probability that they win alternately?

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