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 0 of 2 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 0.0841 . (Round to four decimal places as needed.) (b) The probability that George wins five air hockey games in a row is (Round to four decimal places as needed.)

To find the probabilities for George winning consecutive air hockey games, we need to use the given winning percentage: \( P(\text{win}) = 0.29 \) or \( 29\% \). (a) The probability that George wins two air hockey games in a row is calculated by multiplying the probability of winning one game by itself: \[ P(\text{2 wins}) = P(\text{win}) \times P(\text{win}) = 0.29 \times 0.29 = 0.0841 \] So the probability that George wins two air hockey games in a row is indeed \( 0.0841 \). (b) For the probability that George wins five air hockey games in a row, we multiply the probability of winning one game five times: \[ P(\text{5 wins}) = P(\text{win})^5 = 0.29^5 = 0.00257 \] Thus, rounding to four decimal places, the probability that George wins five air hockey games in a row is \( 0.0026 \). (c) If we want to find the probability that George wins \( 0 \) out of \( 2 \) games, we first determine the probability of losing a game, which is: \[ P(\text{lose}) = 1 - P(\text{win}) = 1 - 0.29 = 0.71 \] Now we can find the probability that George wins \( 0 \) out of \( 2 \) games: \[ P(\text{0 wins out of 2}) = P(\text{lose})^2 = 0.71^2 = 0.5041 \] Next, to calculate the probability that George loses all six games (i.e., does not win six in a row): \[ P(\text{0 wins out of 6}) = P(\text{lose})^6 = 0.71^6 = 0.3623 \] Finally, we want the probability that George wins \( 0 \) out of \( 2 \) games and does not win six in a row. Since these events are independent, we can calculate: \[ P = P(\text{0 wins out of 2}) \times P(\text{not winning 6 in a row}) = 0.5041 \times (1 - 0.3623) \] This leads to the final calculation: \[ P = 0.5041 \times 0.6377 \approx 0.3214 \] In conclusion, the results are: - The probability that George wins five air hockey games in a row is \( 0.0026 \). - The probability that George wins \( 0 \) out of \( 2 \) games but does not win six in a row is approximately \( 0.3214 \).