Q.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?

$P(\text{A wins a game})=3/5$, $P(\text{A loses})=2/5$. $P(\text{exactly 2 of 3})=\binom{3}{2}(\frac{3}{5})^2(\frac{2}{5})=3\times\frac{9}{25}\times\frac{2}{5}=\frac{54}{125}$. $P(\text{all 3})=(\frac{3}{5})^3=\frac{27}{125}$. $P(\text{at least 2})=\frac{54}{125}+\frac{27}{125}=\frac{81}{125}$, so $P(\text{fewer than 2})=1-\fr …

Binomial-style counting over 3 independent games (each game A wins with probability 3/5), summing the 'exactly 2' and 'exactly 3' win cases, then …

Forgetting the binomial coefficient $\binom{3}{2}=3$ when computing 'exactly 2 wins', since there are 3 diffe …