Question

A and B play one another at chess. The probability of A winning a particular game is ⅓ and the 3 probability of a draw is ⅙. In a tournament, A and B play 4 games each other. Find the probability that 6
(i) none is drawn
(ii) A wins 3 games with the other drawn
(iii) B wins exactly 3 games

Answer 1

prob(A wins) = 1/3
prob(A draw) = 1/6
prob(A loses) = 1 - 1/2 - 1/6 = 1/2

Let's look at it as the outcomes of player A
a) no draws:
we could have WWWW = (1/3)^4 = 81
We could have WLLL, in its permutations = 4(1/3)(1/2)^3 = 4/24 = 1/6
We could have WWLL, in its permutations = 4!/(2!2!)(1/3)^2 (1/2)^2 = 1/6
We could have WWWL, in its permutations = 4((1/3)^3 (1/2) = 2/27
prob(no draws) = 1/6 + 1/6 + 2/27 = 11/27

b) must be a permutation of WWWL, which we did above and is 2/27

c) B wins exactly 3 games ----> A loses exactly 3 games
= permutation of WLLL = 4(1/3)(1/2)^3 = 1/6

Your 2 typos could change my interpretation of the question.

Answer 2

for a)
should say:
we could have WWWW = (1/3)^4 = 1/81
...
prob(no draws) = 1/81 + 1/6 + 1/6 + 2/27 = 34/81