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Question
The odds in favour of A winning a game of chess against B are 3:2. If three games are to be played, what are the odds in favour of A's winning at least two games out of the three?
Solution
Let event A: A wins the game and
event B: B wins the game.
Since the odds in favour of A winning a game against B are 3 : 2, the probability of occurrence of event A and B is given by
P(A) = `3/(3 + 2) = 3/5` and P(B) = `2/(3 + 2) = 2/5`
Let event E: A wins at least two games out of three games.
∴ P(E) = P(A) · P(A) · P(B) + P(A) · P(B) · P(A) + P(B) · P(A) · P(A) + P(A) · P(A) · P(A)
= `3/5 xx 3/5 xx 2/5 + 3/5 xx 2/5 xx 3/5 + 2/5 xx 3/5 xx 3/5 + 3/5 xx 3/5 xx 3/5`
= `18/125 + 18/125 + 18/125 + 27/125`
= `81/125`
∴ P(E') = 1 – P(E)
= `1 - 81/125`
= `44/125`
∴ Odds in favour of A’s winning at least two games out of three are
P(E) : P(E') = `81/125 : 44/125`
= 81 : 44
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