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
Solución de inteligencia artificial de Upstudy
Responder
Solución

¡Inicia sesión para desbloquear respuestas gratis!
Una plataforma de aprendizaje en la que confían millones de estudiantes y profesores reales.

Respondido por UpStudy AI y revisado por un tutor profesional
Extra Insights
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}
