On a Saturday evening, 34% of the people in Winnipeg go out to dinner, 18% see a movie, 13% have a party, and 35% stay home. If seven people are randomly selected, what is the probability that one eats
Out, three see a movie, two have a party, and one stays at home?
A) 0.099
B) 0.459
C) 0.102
D) 0.0049

Question

Quizplus · Accepted Answer

The probability of one person going out to dinner is 0.34, so the probability of one going out and the others not going out is: 
(0.34)(0.66)(0.66)(0.66)(0.66)(0.66)(0.66) = (0.34)(0.66)^6

The probability of three people seeing a movie is 0.18^3, and the probability of two people having a party is 0.13^2. The probability of one person staying home is 0.35.

Multiply all of the probabilities together to get:

(0.34)(0.66)^6 * 0.18^3 * 0.13^2 * 0.35 ≈ 0.0049

Therefore, the answer is D.

On a Saturday Evening, 34% of the People in Winnipeg