sir i have a doubt: a and B throw a coin alternately till one of them gets a headand wins the game. Find their respective probabilities of winning.
If he fails the chance of B failing is 1/2 - if that happens the game is back where it started, with A having the same chance of winning as he did the first time he tossed. The chance that A will make a second throw is 1/2 x 1/2 = 1/4
so a trick here is to see that A's chance of winning is
P(A) = 1/2 + (1/4) P(A)
(3/4) P(A) = 1/2
P(A) = 2/3
the first player to toss has a 2/3 chance of winning
If A tosses and fails, probability 1/2, then B is then in the position of the first player to toss, so he has a 2/3 chance of winning, so his chance of winning must be 1/2 x 12/ = 1/3