probability

Question

It has been found that if A and B play a game 12 times, A wins 6 times, B wins 4 times and they draw twice. A and B take part in a series of 3 games. The probability that they will win alternately is?(Alternatively winning means do the draws come in between or it's like ABAB.....so on)

Answer 1

I'd say that "alternately" in three games means either ABA or BAB. So let's find the probably of each case.

 

First let's find the probability of each player winning in one game:

 

The probably of A winning is 6/12 or 1/2.

The probably of B winning is 4/12 or 1/3.

 

So the probability of ABA is 1/2 x 1/3 x 1/2.  (The probabilities are multiplied because each game is a separate independent event.)  That comes to 1/12.

 

The probability of BAB is 1/3 x 1/2 x 1/3 = 1/18.

 

Since the two outcomes are mutually exclusive, then you add the two probabilities together. 1/12 + 1/18 = 5/36.