91Ó°ÊÓ

Teams$A$and $B$ play a series of games, with the first team to win$3$games being declared the winner of the series. Suppose that team $A$ independently wins each game with a probability of $P$.

Let's consider the team $A$has succeeded in the first game.

So, we have$4$games left.

There are several chances.

Team$A$can succeed the series in three games in entire

$(AAA\_scenario)$, or four games in entire($ABAA$_and $A A B A$_scenarios) or in five games in whole ( $ABBAA,ABABA$and $AABBA$scenarios). Therefore, the required probability is

$P(\mathrm{A}\text{wins series}âˆ\pounds \text{A wins the first game})=p^2+2p^2(1−p)+3p^2(1−p{)}^2$

The conditional probability that team A wins the first game is$p^2+2p^2(1−p)+3p^2(1−p{)}^2$.

Teams A and B play a series of games, with the first team to win 3 games being declared the winner of the series. Suppose that team A independently wins each game with probability p.

Apply Baye's theorem,

Here, we have $\frac{P(Awinsthefirstgame|Awinsseries)P(Awinstheseries|Awinsthefirstgame)P(Awinsthefirstgame}{P(Awinsseries)}$Use the same argument in part (a).

Team A can win the series in three, four or five games.

Therefore, $P\left(\text{A wins series}\right)=p^3+3â‹\dots p^3(1−p)+\left(\begin{array}{l}4\\ 2\end{array}\right)p^3(1−p{)}^2$

So, the required probability is$\frac{\left(p^2+2p^2(1−p)+3p^2(1−p{)}^2\right)â‹\dots p}{p^3+3â‹\dots p^3(1−p)+\left(\begin{array}{l}4\\ 2\end{array}\right)p^3(1−p{)}^2}$

The required probability that team A wins the series is$\frac{\left(p^2+2p^2(1−p)+3p^2(1−p{)}^2\right)â‹\dots p}{p^3+3â‹\dots p^3(1−p)+\left(\begin{array}{l}4\\ 2\end{array}\right)p^3(1−p{)}^2}$