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Chapter 3: Problem 78

\(A\) and \(B\) play a series of games. Each game is independently won by \(A\) with probability \(p\) and by \(B\) with probability \(1-p .\) They stop when the total number of wins of one of the players is two greater than that of the other player. The player with the greater number of total wins is declared the winner of the series. (a) Find the probability that a total of 4 games are played. (b) Find the probability that \(A\) is the winner of the series.

Short Answer

Expert verified
The probability of a total of 4 games played is \(2p^2(1-p)\), and the probability that A is the winner of the series is \(3p^2(1-p)\).

Step by step solution

01

Case 1: AABB

The probability of this outcome is given by p^2 (1-p)^2, as we have two games won by A, and two by B.
02

Case 2: BBAA

The probability of this outcome is given by p^2 (1-p)^2 just like in Case 1, as we have two games won by A, and two by B.
03

Case 3: AABA

In this outcome, A wins the first and third games, and B wins the second game. The probability of this case is p^2(1-p).
04

Case 4: BAAB

Similar to Case 3, A wins two games, and B wins one game. The probability of this case is p^2(1-p).
05

Total Probability for 4 games

Adding the probabilities of all four cases, we get the total probability for 4 games: \(2p^2(1-p)^2 + 2p^2(1-p) = 2p^2(1-p)(1 + (1-p)) = 2p^2(1-p) \) (b) Probability that A is the winner of the series
06

Case 1: A wins in 3 games

A must win the first two games in a row, and then B can win the third game. The probability of this case is: \(p^2(1-p)\).
07

Case 2: A wins in 4+ games

We have already calculated the probability of a total of 4 games played: \(2p^2(1-p)\). Among these, A is the winner in cases AABA and BAAB. The probability of A winning in 4 games is: \(2p^2(1-p)\)
08

Total Probability for A to win the series

Adding the probabilities of both cases, we get the total probability of A winning the series: \(p^2(1-p) + 2p^2(1-p) = 3p^2(1-p)\)

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