Question

it has been found that if A B play a game 12 times, A wins 6 times , B wins 4 timesand they draw twice.A and B take part in a series of 3 games. find probabilty that they win alternatively.

Answer 1

As in 12 games , A wins 6 times , so probability of A wins = 6/12 = 1/2

And B wins four times , it means B wins probability = 4/12 = 1/3

Suppose there are three games , then there are two cases in which both wins alternatively.
ABA and BAB , here A and B denotes there winning .

So probability of alternative winning = (1/2* 1/3* 1/2 ) + (1/3 * 1/2 *1/3 ) = 5/36 (ans)