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)` 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