Two players, A and B , alternately draw a ball from a bag containing n red balls and 6 black balls, with replacement. Player A wins the game if he draws a red ball, and player B wins if he draws a black ball. If A makes the first draw and the probability that A wins the game is 4 13 , then the value of n is

Step-by-step solution

Let p be the probability of drawing a red ball. Then p = n n+6 . The probability of drawing a black ball is 1 - p = 6 n+6 . Player A wins on his turn if he draws a red ball, so P(A) = p and P( A ) = 1 - p . Player B wins on his turn if he draws a black ball, so P(B) = 1 - p and P( B ) = 1 - (1 - p) = p . The probability that A wins the game is the sum of the probabilities of A winning on his 1st, 2nd, 3rd turn, etc.: P(A wins ) = P(A) + P( A )P( B )P(A) + P( A )P( B )P( A )P( B )P(A) + This is an infinite geometric