Suppose 20% of the restaurants in a certain part of a town are in violation of t

Question

Suppose 20% of the restaurants in a certain part of a town are in violation of the health code. A health inspector randomly selects eight of the restaurants for inspection. (Round your answers to four decimal places.) (a) What is the probability that none of the restaurants are in violation of the health code? (b) What is the probability that one of the restaurants is in violation of the health code? (c) What is the probability that at least two of the restaurants are in violation of the health code?

Explanation / Answer

Binomial distribution

n = 8

p = 0.2

q = 1 - p = 0.8

P(X) = nCx px qn-x

a) P(none of the restaurants are in viloation of health code)

= P(0)

= 0.88

= 0.1678

b) P(one of the restaurants are in viloation of health code) = P(1)

= 8 x 0.2 x 0.87

= 0.3355

c) P(at least 2 of the restaurants are in viloation of health code) = 1 - P(X < 2)

= 1 - P(0) - P(1)

= 1 - 0.1678 - 0.3355

= 0.4967

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