91Ó°ÊÓ

Using the binomial probability formula, we find the probability of exactly one engine failure (\(P(1)\)): \[P(1) = \binom{4}{1} (0.001)^{1} (1-0.001)^{4-1}\] Calculate the binomial coefficient: \[\binom{4}{1} = \frac{4!}{1!(4-1)!} = 4\] Now, substitute this value back into the formula and compute the probability: \[P(1) = 4 (0.001)^{1} (0.999)^{3}\] \[P(1) \approx 0.00398\]

Using the binomial probability formula, we find the probability of exactly two engine failures (\(P(2)\)): \[P(2) = \binom{4}{2} (0.001)^{2} (1-0.001)^{4-2}\] Calculate the binomial coefficient: \[\binom{4}{2} = \frac{4!}{2!(4-2)!} = 6\] Now, substitute this value back into the formula and compute the probability: \[P(2) = 6 (0.001)^{2} (0.999)^{2}\] \[P(2) \approx 0.00000599\]

Since there are only four engines, more than two engine failures means 3 or 4 engine failures. We'll use the binomial probability formula to find the probabilities for these scenarios and then add them together. Calculate the probability for exactly three engine failures (\(P(3)\)): \[P(3) = \binom{4}{3} (0.001)^{3} (1-0.001)^{4-3}\] Calculate the binomial coefficient: \[\binom{4}{3} = \frac{4!}{3!(4-3)!} = 4\] Now, substitute this value back into the formula and compute the probability: \[P(3) = 4 (0.001)^{3} (0.999)^{1}\] \[P(3) \approx 0.000000003996\] Calculate the probability for exactly four engine failures (\(P(4)\)): \[P(4) = \binom{4}{4} (0.001)^{4} (1-0.001)^{4-4}\] Calculate the binomial coefficient: \[\binom{4}{4} = \frac{4!}{4!(4-4)!} = 1\] Now, substitute this value back into the formula and compute the probability: \[P(4) = 1 (0.001)^{4} (0.999)^{0}\] \[P(4) \approx 0.000000000001\] Now, add the probabilities for 3 and 4 engine failures to find the probability of more than two engine failures: \[P(\text{more than 2}) = P(3) + P(4) \approx 0.000000003996 + 0.000000000001\] \[P(\text{more than 2}) \approx 0.000000003997\] #Answer# a. The probability of exactly one engine failure is approximately 0.00398. b. The probability of exactly two engine failures is approximately 0.00000599. c. The probability of more than two engine failures (and, therefore, a crash) is approximately 0.000000003997.