From experience, an airline knows that only \( 85 \% \) of the passengers booked for a certain flight actually show up. If 9 passengers are randomly selected, find the probability that at most 6 of them show up. Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places. (If necessary, consult a list of formulas.)

To solve the problem, we need to model it using the binomial distribution, where \( n = 9 \) (the number of selected passengers) and \( p = 0.85 \) (the probability that a passenger shows up). We want to find the probability that at most 6 passengers show up, which means we need to calculate the probability for \( X \) where \( X \) is the number of passengers that show up, and we want \( P(X \leq 6) \). Using the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( \binom{n}{k} \) is the binomial coefficient. We need to calculate: \[ P(X \leq 6) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) \] ### Calculations 1. **Compute individual probabilities**: - \( P(X = 0) = \binom{9}{0} (0.85)^0 (0.15)^9 = 1 \cdot 1 \cdot 0.00019683 = 0.00019683 \) - \( P(X = 1) = \binom{9}{1} (0.85)^1 (0.15)^8 = 9 \cdot 0.85 \cdot 0.00003051 \approx 0.00023345 \) - \( P(X = 2) = \binom{9}{2} (0.85)^2 (0.15)^7 = 36 \cdot 0.7225 \cdot 0.000004576 \approx 0.00106963 \) - \( P(X = 3) = \binom{9}{3} (0.85)^3 (0.15)^6 = 84 \cdot 0.614125 \cdot 0.0000006875 \approx 0.00488372 \) - \( P(X = 4) = \binom{9}{4} (0.85)^4 (0.15)^5 = 126 \cdot 0.52200625 \cdot 0.0000001024 \approx 0.00630381 \) - \( P(X = 5) = \binom{9}{5} (0.85)^5 (0.15)^4 = 126 \cdot 0.4437053125 \cdot 0.000017576 \approx 0.06787658 \) - \( P(X = 6) = \binom{9}{6} (0.85)^6 (0.15)^3 = 84 \cdot 0.376389578125 \cdot 0.0003375 \approx 0.10647347 \) 2. **Sum up the probabilities**: \[ P(X \leq 6) = 0.00019683 + 0.00023345 + 0.00106963 + 0.00488372 + 0.00630381 + 0.06787658 + 0.10647347 \approx 0.186345 \] Finally, rounding to two decimal places, we have: \[ P(X \leq 6) \approx 0.19 \] Therefore, the probability that at most 6 passengers show up is **0.19**.