Answer to Question #267797 in Statistics and Probability for cjk

Question

. A shipment of 15 concrete piles has been received by a contractor to be used in a bridged project. Suppose that 6 of these cylinders have crushing strength below the specified minimum, if 5 cylinders are randomly chosen, what is the probability that at least 2 concrete cylinder will have crushing strength below minimum?

Answer 1

N=15

n=5

k=6

Hypergeometric distribution

$P(X=x) = \frac{\binom{k}{x} \binom{N-k}{n-x}}{\binom{N}{n}} \\ P(X≥2) = 1 -[P(X=0) + P(X=1)] \\ = 1 -[(\frac{\binom{5}{0} \binom{15-6}{5-0}}{\binom{15}{5}}) + (\frac{\binom{5}{1} \binom{15-6}{5-1}}{\binom{15}{5}})] \\ = 1 -[0.0419 + 0.2098] \\ = 0.7483$