91Ó°ÊÓ

Team A and B play a sequence of games against each other in the world series of baseball.

Team A is the winner by playing total 4 games.

Let p be the probability that team A will win any particular game.

If a team A wins only three matches in 6 games, then it will be necessary to play the seventh game to win the series.

Hence to win the series in seven games it is necessary to win 3 matches in 6 games.

Let X be the number of wins.

Total number of games played be \(n = 6\)

Hence X follows binomial distribution with parameters \(n = 6\) and p.

A probability mass function of X is given as

\({\rm P}\left( {X = x} \right) = \left( \begin{aligned}{l}n\\x\end{aligned} \right){p^x}{\left( {1 - p} \right)^{n - x}}\).

Hence, the probability that it will be necessary to play seven games to win the series is given as:

\({\rm P}\left( {X = x} \right) = \left( \begin{aligned}{l}6\\3\end{aligned} \right){p^3}{\left( {1 - p} \right)^3}\).

\(p = \frac{1}{3}\)

Therefore,

\(\begin{aligned}{}{\rm P}\left( {X = 3} \right) = \left( \begin{aligned}{l}6\\3\end{aligned} \right){\left( {\frac{1}{3}} \right)^3}{\left( {1 - \frac{1}{3}} \right)^3}\\ = 20 \times \frac{1}{{27}} \times \frac{8}{{27}}\\ = 0.2194\end{aligned}\)

Therefore the probability that it will be necessary to play seven games to win the series is 0.2194.