Question

Team A has played with team B 15 times during the season. team A has won 10 times and Team B 5 times. if we take the probability that A will win a game against B to be 2/3. find the probability that a will win against B -

a) the next 3 games

b) 2 of the next 3 games

c) at least 2 of the next 3 games

Answer 1

P(A wins 1 game)=$\frac{2}{3}$

So,

(a) P(A wins 3 games)=$\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3}=\frac{8}{27}$

(b) P(A wins 2 out of 3 games)$$\begin{gathered} =\mathrm{P}(\mathrm{A}\mathrm{wins}2\mathrm{games}\mathrm{and}\mathrm{loses}1\mathrm{game}) \\ =\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}=\frac{4}{27} \end{gathered}$$

(c) P(A wins at least 2 out of 3 games)=$$\begin{gathered} =\mathrm{P}(\mathrm{A}\mathrm{wins}2\mathrm{games})+\mathrm{P}(\mathrm{A}\mathrm{wins}3\mathrm{games}) \\ =\frac{2}{3}\times \frac{2}{3}\times \frac{1}{3}+\frac{2}{3}\times \frac{2}{3}\times \frac{2}{3} \\ =\frac{4}{27}+\frac{8}{27}=\frac{4}{9} \end{gathered}$$