Pregunta
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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

Ask by Sanders Martin. in the United States
Mar 15,2025

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The probability that exactly 6 flights are on time is approximately 0.226. The probability that at most 3 flights are on time is approximately 0.113. The probability that at least 7 flights are on time is approximately 0.274.

Solución

**Step 1. Define the problem using the binomial distribution** We have a binomial random variable \( X \) representing the number of on‐time flights out of 11 trials with probability \( p=0.50 \). Therefore, \[ X \sim \text{Binomial}(n=11, p=0.50). \] The probability mass function is given by \[ P(X=k)=\binom{11}{k}(0.50)^k(0.50)^{11-k}=\binom{11}{k}\left(0.50\right)^{11}. \] **Step 2. Compute the probability that exactly 6 flights are on time** We need \[ P(X=6)=\binom{11}{6}\left(0.50\right)^{11}. \] Calculate the binomial coefficient: \[ \binom{11}{6} = \frac{11!}{6!5!} = 462. \] Thus, \[ P(X=6)=462\left(0.50\right)^{11} = \frac{462}{2048}\approx 0.2256. \] Rounded to 3 significant figures, this is approximately \[ 0.226. \] **Step 3. Compute the probability that at most 3 flights are on time** We need the cumulative probability for \( k=0 \) to \( k=3 \): \[ P(X\le3)=\sum_{k=0}^{3}\binom{11}{k}\left(0.50\right)^{11}. \] Calculate each term: - For \( k=0 \): \(\binom{11}{0}=1\) - For \( k=1 \): \(\binom{11}{1}=11\) - For \( k=2 \): \(\binom{11}{2}=55\) - For \( k=3 \): \(\binom{11}{3}=165\) Summing the coefficients: \[ 1+11+55+165=232. \] Thus, \[ P(X\le3)=232\left(0.50\right)^{11}=\frac{232}{2048}\approx 0.1133. \] Rounded to 3 significant figures, this is approximately \[ 0.113. \] **Step 4. Compute the probability that at least 7 flights are on time** We need the probability for \( k \ge 7 \): \[ P(X\ge7)=\sum_{k=7}^{11}\binom{11}{k}\left(0.50\right)^{11}. \] Calculate the required binomial coefficients: - For \( k=7 \): \(\binom{11}{7} = 330\) - For \( k=8 \): \(\binom{11}{8} = 165\) - For \( k=9 \): \(\binom{11}{9} = 55\) - For \( k=10 \): \(\binom{11}{10} = 11\) - For \( k=11 \): \(\binom{11}{11} = 1\) Sum the coefficients: \[ 330 + 165 + 55 + 11 + 1 = 562. \] Thus, \[ P(X\ge7)=562\left(0.50\right)^{11}=\frac{562}{2048}\approx 0.2744. \] Rounded to 3 significant figures, this is approximately \[ 0.274. \] **Final Answers:** 1. The probability that exactly 6 flights are on time is approximately \( 0.226 \). 2. The probability that at most 3 flights are on time is approximately \( 0.113 \). 3. The probability that at least 7 flights are on time is approximately \( 0.274 \).

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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}

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