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Question

When commercial aircraft are inspected, wing cracks are reported as nonexistent, detectable, or critical. The history of a particular fleet indicates that 70% of the planes inspected have no wing cracks, 25% have detectable wing cracks, and 5% have critical wing cracks. Five planes are randomly selected. Find the probability that one has a critical crack, two have detectable cracks, and two have no cracks.

Solution

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Definition multinomial distribution:

p(y1,y2,...,yk)=n!y1!y2!...yk!p1y1p2y2...pkykp(y_1,y_2,...,y_k)=\dfrac{n!}{y_1!y_2!...y_k!}p_1^{y_1}p_2^{y_2}...p_k^{y_k}

Then we obtain:

p(2,2,1)=5!2!2!1!0.7020.2520.0510.0459p(2,2,1)=\dfrac{5!}{2!2!1!}0.70^{2} 0.25^2 0.05^1\approx 0.0459

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