Answer to Question #226054 in Statistics and Probability for ahmed

Question

Identical computer components are shipped in boxes of 5. About 15% of components

have defects. Boxes are tested in a random order with replacement.

a. What is the probability that a randomly selected box has only non-defective

components?

b. What is the probability that at least 8 of randomly selected 10 boxes have only

non-defective components?

Answer 1

X=the number of defective components

X follows a binomial distribution with n=5 and p=0.15

$P(X=x) = C^n_xp^x(1-p)^{n-x}$

a.

$P(X=0) = \frac{5!}{0!(5-0)!} \times 0.15^0 \times (0.85)^{5-0} \\ = 1 \times 1 \times 0.4437 \\ = 0.4437$

b. X=the number of non-defective components

X follows a binomial distribution with $n=10 \times 5 = 50$ and p=0.85

At least 8 of randomly selected 10 boxes will have 40, 45 or 50 non-defective components.

$P = P(X=40) + P(X=45) + P(X=50) \\ P(X=40) = \frac{50!}{40!(50-40)!} \times 0.85^{40} \times 0.15^{50-40} \\ = 209638330 \times 0.0015023 \times 5.7665 \times 10^{-9} \\ = 0.001816 \\ P(X=45) = \frac{50!}{45!(50-45)!} \times 0.85^45 \times 0.15^{50-45} \\ = 1945800 \times 0.000666 \times 7.5937 \times 10^{-5} \\ = 0.098493 \\ P(X=50) = \frac{50!}{50!(50-50)!} \times 0.85^50 \times 0.15^{50-50} \\ = 1 \times 0.1968 \times 1 \\ = 0.000295 \\ P = 0.001816 + 0.098493 + 0.000295= 0.100604$