Question

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. Find the conditional probability that team A wins

(a) the series given that it wins the first game;

(b) the first game given that it wins the series.

Short answer

1. The probability that team A wins the first game is $P(\text{A wins series}\mid \text{A wins the first game})=p^2+2p^2(1-p)+3p^2(1-p{)}^2$
2. The probability that team A wins the series is$P\left(\text{A wins series}\right)=p^3+3\cdot p^3(1-p)+\left(\begin{array}{l}4\\ 2\end{array}\right)p^3(1-p{)}^2$

Step-by-step solution

Step 1:Given Information (Part a)

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 $P$.

Step 2:Explanation (part a)

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 $A\hspace{0.17em}A\hspace{0.17em}B\hspace{0.17em}A$_scenarios) or in five games in whole ( $ABBAA,ABABA$and $AABBA$scenarios). Therefore, the required probability is

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

Final Answer (Part a)

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

Given Information (Part b)

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.

Solution (Part b)

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, $P\left(\text{A wins series}\right)=p^3+3\cdot p^3(1-p)+\left(\begin{array}{l}4\\ 2\end{array}\right)p^3(1-p{)}^2$

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

Final Answer 

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