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?

Answer 1

Probability of winning of A is
$p(A) = \frac{3}{5}$
Probability of winning B is
$p(B) = \frac{2}{5}$
Probability of winning of a at least two times

$p(A)p(A)p(A) + p(A)p(A)p(B) \\ \\ = \frac{3}{5} \times \frac{3}{5} \times \frac{3}{5} + \frac{3}{5} \times \frac{3}{5} \times \frac{2}{5} \\ \\ = \frac{27}{125} + \frac{18}{125} \\ \\p(W) = \frac{45}{125} \\ \\= \frac{9}{25}$
odds in favour of winning of A at least two times

$\frac{p(W)}{1 - p(W)} \\ \\ = \frac{ \frac{9}{25} }{1 - \frac{9}{25} } \\ \\ = \frac{9}{16}$
odds in favour of winning A at least twice 9:16

Hope it helps you