Suppose Emerson wins $25 \%$ of all air hockey games.
(a) What is the probability that Emerson wins two air hockey games in a row?
(b) What is the probability that Emerson wins three air hockey games in a row?
(c) When events are independent, their complements are independent as well. Use this result to determine the probability that Emerson wins three air hockey games in a row, but does not win four in a row.
(a) The probability that Emerson wins two air hockey games in a row is 0.0625 .
(Round to four decimal places as needed)
(b) The probability that Emerson wins three air hockey games in a row is 0.0156 '
(Round to four decimal places as needed)
(c) The probability that Emerson wins three air hockey games in a row, but does not win four in a row is (Round to four decimal places as needed.)
Answer
Final Answer: (a) The probability that Emerson wins two air hockey games in a row is \(\boxed{0.0625}\). (b) The probability that Emerson wins three air hockey games in a row is \(\boxed{0.0156}\). (c) The probability that Emerson wins three air hockey games in a row, but does not win four in a row is \(\boxed{0.0117}\).
Step 1 ：Given that Emerson wins 25% of all air hockey games, we are asked to find the probability of Emerson winning consecutive games.
Step 2 ：Since the games are independent events, the probability of multiple events occurring is the product of their individual probabilities.
Step 3 ：For part (a), we need to find the probability that Emerson wins two games in a row. Since the probability of winning a single game is 0.25, the probability of winning two games in a row is \(0.25 \times 0.25 = 0.0625\).
Step 4 ：For part (b), we need to find the probability that Emerson wins three games in a row. This is \(0.25 \times 0.25 \times 0.25 = 0.015625\).
Step 5 ：For part (c), we need to find the probability that Emerson wins three games in a row but does not win the fourth game. This is the probability of winning three games in a row (\(0.25 \times 0.25 \times 0.25\)) multiplied by the probability of losing the fourth game (\(1 - 0.25 = 0.75\)). So, the probability is \(0.015625 \times 0.75 = 0.01171875\).
Step 6 ：Final Answer: (a) The probability that Emerson wins two air hockey games in a row is \(\boxed{0.0625}\). (b) The probability that Emerson wins three air hockey games in a row is \(\boxed{0.0156}\). (c) The probability that Emerson wins three air hockey games in a row, but does not win four in a row is \(\boxed{0.0117}\).
