Question

A and B play 12 games of chess, which 6 are won by A, 4 are won by B and 2 end in a draw. They agree to play a match consisting of 3 games. Based on the past outcomes what is the probability that B wins at least 2 games?

Answer 1

Solution:
Total number game of chess = 12
P(Win A) = 6/12 = 1/2
P(Win B) = 4/12 = 1/3
P(Draw) = 2/12 = 1/6
Now they agree to play a match consisting of 3 games, We need to calculate the probability that B wins at least 2 games
Here we will use binomial probability distribution because all games are independent of each other so probability can be calculated as
P(X =n | N,p) = NCn*(p^n)*((1-p)^(N-n))
P(X>=2) = P(X=2) + P(X=3) = 3C2*(1/3)^2*(1-(1/3))^(3-2) + 3C3*(1/3)^3*(1-(1/3))^(3-3) = 3*0.1111*0.6667 + 0.037 = 0.2222 + 0.037 = 0.2593
So there is 25.93% probability that B will wins atleast 2 games.