In a sample of 900 U S adults, 192 dine out at a restaurant more than once per w

Question

In a sample of 900 U S adults, 192 dine out at a restaurant more than once per week. Two U. S. adults are selected at random from the population of all U. S. adults without replacement. Assuming the sample is representative of all U. S. adults, complete parts (a) through (d). (a) Find the probability that both adults dine out more than once per week. The probability that both adults dine out more than once per week is 0 045. (Round to three decimal places as needed.) (b) Find the probability that neither adult dines out more than once per week. The probability that neither adult dines out more than once per week is (Round to three decimal places as needed.)

Explanation / Answer

Solution:-

n = 900, x = 192

p = 192/900

p = 0.2133

a) The probability that both adults dine out more than once per week is 0.0455

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

n = 2

P(x = 2) = 0.0455

b) The probability that neither adults dine out more than once per week is 0.619.

By applying binomial distribution:-

P(x, n, p) = nCx*p x *(1 - p)(n - x)

n = 0

P(x = 0) = 0.619

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