Question: Your flight has been delayed. At Denver International Airport, 85% of recent flights have arrived…

Your flight has been delayed. At Denver International Airport, 85% of recent flights have arrived on time. A sample of 12 flights is studied. Round the probabilities to four decimal places.

Part 1 of 4 (a) Find the probability that all 12 of the flights were on time. The probability that all 12 of the flights were on time is \(\boxed{\phantom{0.0000}}\).

Part 2 of 4 (b) Find the probability that exactly 10 of the flights were on time. The probability that exactly 10 of the flights were on time is \(\boxed{\phantom{0.0000}}\).

Part 3 of 4 (c) Find the probability that 10 or more of the flights were on time. The probability that 10 or more of the flights were on time is \(\boxed{\phantom{0.0000}}\).

Part 4 of 4 (d) Would it be unusual for 11 or more of the flights to be on time? It \(\boxed{\text{(Choose one)}}\) be unusual for 11 or more of the flights to be on time since the probability is \(\boxed{\phantom{0.0000}}\).

Solution

To solve these problems, we can use the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where: - \( n \) is the number of trials (flights in this case), - \( k \) is the number of successful trials, - \( p \) is the probability of success (flights being on time), - \(\binom{n}{k}\) is the binomial coefficient. Given: \( n = 12 \), \( p = 0.85 \) Part 1: Find the probability that all 12 of the flights were on time. \[ P(X = 12) = \binom{12}{12} (0.85)^{12} (1-0.85)^{0} \] Calculate: \[ P(X = 12) = 1 \times (0.85)^{12} \times 1 \] \[ P(X = 12) = (0.85)^{12} \] \[ P(X = 12) \approx 0.1487 \] Part 2: Find the probability that exactly 10 of the flights were on time. \[ P(X = 10) = \binom{12}{10} (0.85)^{10} (1-0.85)^{12-10} \] Calculate: \[ P(X = 10) = \frac{12 \times 11}{2 \times 1} \times (0.85)^{10} \times (0.15)^2 \] \[ P(X = 10) = 66 \times (0.85)^{10} \times (0.15)^2 \] \[ P(X = 10) \approx 0.2310 \] Part 3: Find the probability that 10 or more of the flights were on time. \[ P(X \geq 10) = P(X = 10) + P(X = 11) + P(X = 12) \] First, calculate \( P(X = 11) \): \[ P(X = 11) = \binom{12}{11} (0.85)^{11} (0.15)^1 \] \[ P(X = 11) = 12 \times (0.85)^{11} \times 0.15 \] \[ P(X = 11) \approx 0.3109 \] Now add the probabilities: \[ P(X \geq 10) = 0.2310 + 0.3109 + 0.1487 \] \[ P(X \geq 10) \approx 0.6906 \] Part 4: Would it be unusual for 11 or more of the flights to be on time? Calculate \( P(X \geq 11) \): \[ P(X \geq 11) = P(X = 11) + P(X = 12) \] \[ P(X \geq 11) = 0.3109 + 0.1487 \] \[ P(X \geq 11) \approx 0.4596 \] Since this probability is not very low, it would not be considered unusual for 11 or more flights to be on time.