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Suppose that a game-playing process continues until a specific event (team A or B wins \(i\) games) occurs. A geometric distribution models the number of trials required for the first successful event (winning \(i\) games) to occur. The probability mass function for a geometric random variable X is given by: \(P(X = k) = p(1-p)^{k-1}\), where k is the number of trials required and p is the probability of success (in this case, team A winning a game).

The expected value for the geometric random variable can be calculated by the formula: \(E[X]=\sum_{k=1}^{\infty} kP(X=k) = \sum_{k=1}^{\infty} kp(1-p)^{k-1}\). For our problem, the sum has to be modified to include only scenarios where team A or B has won \(i\) games. (a) \(i=2\):

For the game series to end with either team A or B winning 2 games, the possible lengths of the series are 2, 3, or 4 games. Let's find the probabilities for each of these scenarios and calculate their weighted sum. - Length 2: The probability that team A wins both games is \(p^2\). The probability that team B wins both games is \((1-p)^2\). Thus, the probability of the series ending with two games is \(p^2+(1-p)^2\). The expected number of games for this scenario is \(2(p^2+(1-p)^2)\). - Length 3: The probability that team A wins two games and then team B wins one game is \(p^2(1-p)\), and vice versa, is \((1-p)^2p\). Thus, the probability of the series ending with three games is \(2p^2(1-p)\). The expected number of games for this scenario is \(3(2p^2(1-p))\). - Length 4: The probability that team A wins two games, then team B wins two games is \(p^2(1-p)^2\), and vice versa, is \((1-p)^2p^2\). Thus, the probability of the series ending with four games is \(2p^2(1-p)^2\). The expected number of games for this scenario is \(4(2p^2(1-p)^2)\). The total expected number of games for the case when \(i=2\) is given by the sum of these three expected values: \(E[X] = 2(p^2+(1-p)^2) + 3(2p^2(1-p)) + 4(2p^2(1-p)^2)\). (b) \(i=3\):

Repeat the previous calculations for the case when \(i=3\). Then analyze the sums to determine whether the number of games played is maximized when \(p=\frac{1}{2}\). Finally, by evaluating the given expressions for the expected number of games and showing that the maximum value occurs when \(p=\frac{1}{2}\), we have solved both the cases for \(i=2\) and \(i=3\).