Score: 0 of 4 points United Airlines' flights from Chicago to Seattle are on time \( 50 \% \) of the time. Suppose 11 flights are randomly selected, and the number on-time flights is recorded. (Round answers to 3 significant figures) The probability that exactly 6 flights are on time is Enter your answer The probability that at most 3 flights are on time is Enter your answer The probability that at least 7 flights are on time is Enter your answer

To solve this problem, we can model the situation using the binomial probability formula since we are dealing with a fixed number of trials (11 flights), two possible outcomes (on time or not on time), and a constant probability of success (50% on time). The binomial probability formula is given by: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \(n\) is the total number of trials (flights), - \(k\) is the number of successes (on-time flights), - \(p\) is the probability of success (on-time), - \(\binom{n}{k}\) is "n choose k," the binomial coefficient. In our case: - \(n = 11\), - \(p = 0.5\). Let's compute the required probabilities step-by-step: ### 1. Probability that exactly 6 flights are on time: Using the formula: \[ P(X = 6) = \binom{11}{6} (0.5)^6 (0.5)^{11-6} = \binom{11}{6} (0.5)^{11} \] Calculating \(\binom{11}{6}\): \[ \binom{11}{6} = \frac{11!}{6!(11-6)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462 \] Now substituting back into the probability formula: \[ P(X = 6) = 462 \cdot (0.5)^{11} = 462 \cdot \frac{1}{2048} \approx 0.225 \] ### 2. Probability that at most 3 flights are on time: To find \(P(X \leq 3)\), sum the probabilities for \(k = 0\), \(1\), \(2\), and \(3\): \[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \] Calculating each: - For \(k = 0\): \[ P(X = 0) = \binom{11}{0} (0.5)^{11} = 1 \cdot \frac{1}{2048} \approx 0.000488 \] - For \(k = 1\): \[ P(X = 1) = \binom{11}{1} (0.5)^{11} = 11 \cdot \frac{1}{2048} \approx 0.00537 \] - For \(k = 2\): \[ P(X = 2) = \binom{11}{2} (0.5)^{11} = 55 \cdot \frac{1}{2048} \approx 0.0269 \] - For \(k = 3\): \[ P(X = 3) = \binom{11}{3} (0.5)^{11} = 165 \cdot \frac{1}{2048} \approx 0.0806 \] Now summing these results: \[ P(X \leq 3) = 0.000488 + 0.00537 + 0.0269 + 0.0806 \approx 0.1134 \] ### 3. Probability that at least 7 flights are on time: To find \(P(X \geq 7)\): \[ P(X \geq 7) = P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) \] Calculating each: - For \(k = 7\): \[ P(X = 7) = \binom{11}{7} (0.5)^{11} = 330 \cdot \frac{1}{2048} \approx 0.161 \] - For \(k = 8\): \[ P(X = 8) = \binom{11}{8} (0.5)^{11} = 165 \cdot \frac{1}{2048} \approx 0.0806 \] - For \(k = 9\): \[ P(X = 9) = \binom{11}{9} (0.5)^{11}