Question

In a population of interest, we know that, 77% drink coffee, and 23% drink tea. Assume that drinking coffee and tea are disjoint events in this population. We also know coffee drinkers have a 30% chance of smoking. There is a 13% chance of smoking for those who drink tea.

a. (2 points) Five individuals are randomly chosen from this population. What is the probability that four of them drink coffee?

b. (2 points) Five individuals are randomly chosen from this population. What is the probability that the first four drink coffee and the last one drinks tea?

c. (2 points) Five individuals are randomly chosen from this population. What is the expected number (population mean) of coffee drinkers?

d. (2 points) Five individuals are randomly chosen from this population. Find the standard deviation for the number of coffee drinkers.

e. (3 points) A person is randomly chosen from this population. What is the probability that the person smokes?

f. (3 points) We know that a selected person is a smoker, what is the probability that the person drinks coffee?

Answer 1

a) Here, n = 5

p = 0.77

It is a binomial distribution.

P(X = x ) = nCx * px * (1 - p)n - x

P(X = 4) = 5C4 * (0.77)^4 * (0.23)^1 = 0.4043

b) Probability = (0.77)^4 * 0.23 = 0.0809

c) n = 5

p = 0.77

Expected value = np = 5 * 0.77 = 3.85

d) Standard deviation = sqrt(np(1 - p))

= sqrt(5 * 0.77 * 0.23) = 0.9410

e) P(smokes) = P(smokes| drinks coffee) * P(drinks coffee) + P(smokes|drinks tea) * P(drinks tea)

= 0.3 * 0.77 + 0.13 * 0.23 = 0.2609

f) P(drinks coffee|smokes) = P(smokes|drinks coffee) * P(drinks coffee)/P(smokes) = (0.3 * 0.77)/0.2609 = 0.8854