Two evenly matched basketball teams (call them A and B) compete in a best-of 7 championships (the first team to win 4 games wins the championship). Once the champion has been determined, no more games are played. In each game, there is a home team and an away team. The home team wins the game with probability p ≥ 1/2, independent of all previous games. Suppose that the first three games will be held at the home of team A and the last 4 (or fewer if they are not needed) are played at the home of team B.(a) Let X be the number of games won by team A out of the first 3 games. Specify the distribution of X.(b) Find the probability that only 4 games are played.(c) Which of the two teams is more likely to win the trophy? Explain why.(d) Give an expression for the probability that team A wins the trophy, and evaluate it when p = 0.55.(e) Let Y be the number of games won by team A. Find the probability mass function for Y.(f) Evaluate the expected number of games won by team A and the expected number of games played when p = 1/2.(g) Observe (via computations or simulation) that when p = 0.55 the expected number of games won by team A is larger than that of team B, even though team B is more likely to win the trophy.

(a) X follows a binomial distribution with parameters n=3 (the number of trials, which is the number of games played at team A's home) and p (the probability of success, which is the probability that team A wins a game at home).

(b) The probability that only 4 games are played is the probability that team A wins the first 3 games and team B wins the 4th game. This is given by p^3 * (1-p).

(c) Team B is more likely to win the trophy because they have the home advantage in more games (4 games compared to 3 for team A).

(d) The probability that team A wins the trophy can be calculated by considering all the possible ways team A can win 4 games. This is given by the sum of the probabilities of the following events: team A wins the first 4 games, team A wins 4 out of the first 5 games, team A wins 4 out of the first 6 games, and team A wins 4 out of 7 games. When p=0.55, this probability can be calculated using the binomial distribution formula.

(e) Y also follows a binomial distribution with parameters n=7 (the total number of games) and p (the probability that team A wins a game).

(f) When p=1/2, the expected number of games won by team A is np = 7/2 = 3.5. The expected number of games played is the sum of the probabilities of each possible number of games multiplied by the number of games, which can be calculated using the formula for the expected value of a random variable.

(g) When p=0.55, the expected number of games won by team A is np = 7*0.55 = 3.85, which is larger than the expected number of games won by team B (3.15). However, team B is more likely to win the trophy because they have the home advantage in more games. This can be confirmed by calculating the probability that each team wins the trophy.

(a) X follows a binomial distribution with parameters n=3 (the number of trials, which is the number of games played at team A's home) and p (the probability of success, which is the probability that team A wins a game at home).

(b) The probability that only 4 games are played is the probability that team A wins the first 3 games and team B wins the 4th game. This is given by p^3 * (1-p).

(c) Team B is more likely to win the trophy because they have the home advantage in more games (4 games compared to 3 for team A).

(d) The probability that team A wins the trophy can be calculated by considering all the possible ways team A can win 4 games. This is a complex calculation that involves binomial coefficients and powers of p and (1-p). When p=0.55, this calculation gives a probability of approximately 0.487.

(e) Y also follows a binomial distribution, but with parameters n=7 (the total number of games) and p (the probability that team A wins a game).

(f) When p=1/2, the expected number of games won by team A is np = 7/2 = 3.5. The expected number of games played is a more complex calculation, but it can be shown to be approximately 5.81.

(g) When p=0.55, the expected number of games won by team A is np = 7*0.55 = 3.85, which is larger than the expected number of games won by team B (3.15). However, as explained in part (c), team B is more likely to win the trophy because they have the home advantage in more games.