Team A has probability 2/3 of winning whenever it plays. If A plays 6 games, find the probability that A wins, 1. two games, 2. at least one game, 3. more than half of its games.

The is binomial in nature.

p=2/3, q=1/3

part 1: C(6,1)(\frac{2}{3}^{1}(\frac{1}{3})^{5}

part 2, same as above only with 2 instead of 1.

part 3: The best thing for at least 1 is find the probability of 0 and subtract from 1:

1-\left[C(6,0)(\frac{2}{3})^{0}(\frac{1}{3})^{6}\right]

part 4: More than half means 4, 5, or 6 games.

\sum_{k=4}^{6}C(6,k)(\frac{2}{3})^{k}(\frac{1}{3}) ^{6-k}

I will leave you do the number crunching. Okey-doke?