Suppose George wins \( 29 \% \) of all air hockey games. (a) What is the probability that George wins two air hockey games in a row? (b) What is the probability that George wins five 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 George wins five air hockey games in a row, but does not win six in a row. (a) The probability that George wins two air hockey games in a row is (Round to four decimal places as needed.)

To find the probability that George wins two air hockey games in a row, we multiply the probabilities of him winning each game. Since he wins \(29\%\) of the time, that probability is \(0.29\). (a) The probability that George wins two air hockey games in a row is: \( P(\text{win 2 in a row}) = P(\text{win 1}) \times P(\text{win 2}) = 0.29 \times 0.29 = 0.0841 \). (b) For George to win five air hockey games in a row, we again multiply the probabilities together: \( P(\text{win 5 in a row}) = P(\text{win 1})^5 = 0.29^5 = 0.0023 \) (approximately). (c) To find the probability that George wins five games in a row but does not win six in a row, we first say that the probability of losing the sixth game is \(1 - P(\text{win}) = 0.71\). Thus, \( P(\text{win 5 and not win 6}) = P(\text{win 5}) \times P(\text{lose 6}) = 0.0023 \times 0.71 \approx 0.0016 \). And voilà! The answers you've been looking for!