Q.4.11 - A First Course in Probability

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. 

Step-by-Step Solution

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 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

$( A A A \_scenario)$, or four games in entire($A B A A$ _and $A\hspace{0.17em}A\hspace{0.17em}B\hspace{0.17em}A$_scenarios) or in five games in whole ( $A B B A A, A B A B A$ and $A A B B A$ scenarios). Therefore, the required probability is

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

Step 3: Final Answer (Part a)

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

Step 4: 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. 

Step 5: Solution (Part b)

Apply Baye's theorem, 

Here, we have $\frac{P(A wins the first game| A wins series) P(A wins the series|A wins the first game)P(A wins the first game}{P(A wins series)}$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 $\frac{\left(p^2+2p^2(1-p)+3p^2(1-p{)}^2\right)\cdot p}{p^3+3\cdot p^3(1-p)+\left(\begin{array}{l}4\\ 2\end{array}\right)p^3(1-p{)}^2}$

Step 6: Final Answer

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