Answer:
To find the probability for each scenario, we can use combinations and the probability of winning a single game, assuming each player has an equal chance of winning. Since there are only two players, there are two possible outcomes for each game: a win for Ryan or a win for Alex.
i) Ryan wins 3 games:
There are a total of 3 games to be played, and Ryan needs to win 3 of them. The number of ways to choose which games Ryan wins out of the 3 is given by the combination formula: C(3,3) = 1. The probability of Ryan winning a single game is 1/2, so the probability of Ryan winning all 3 games is (1/2)^3 = 1/8. Therefore, the probability of Ryan winning exactly 3 games in the next set of 3 is:
P(Ryan wins 3 games) = C(3,3) * (1/8) = 1/8
ii) Two games end in a draw:
There is only 1 way for 2 games out of 3 to end in a draw (assuming no ties in the third set). The probability of a draw in a single game is also 1/2. So, the probability of two games ending in a draw is (1/2)^2 = 1/4. Therefore, the probability of two games ending in a draw in the next set of 3 is:
P(Two games end in a draw) = 1 * (1/4) = 1/4
iii) Each player wins alternatively:
There are 2 ways that the players can alternate wins in a 3-game set: either Ryan wins the first and third games, or Alex wins the first and third games. In each case, the probability is (1/2) * (1/2) = 1/4. Therefore, the probability of each player winning alternatively in the next set of 3 is:
P(Each player wins alternatively) = 2 * (1/4) = 1/2
iv) Alex wins at least one game:
The only way that Alex cannot win any games is if Ryan wins all 3 games, which we calculated in part i to have a probability of 1/8. Therefore, the probability of Alex winning at least one game in the next set of 3 is:
P(Alex wins at least one game) = 1 - P(Ryan wins