Probability and statistics

Question

On one large campus, 18% of students surveyed said that they spend less than an hour a night studying. If three students are chosen randomly, find the following probabilities.

Part 1 of 2

(a) What's the probability that all three spend less than an hour a night studying? Express your answer as a decimal rounded to four decimal places.

The probability that all three spend less than an hour a night studying is

Part 2 of 2

(b) What's the probability that two of the three spend less than an hour studing? Express vour answer as a decimal rounded to four decimal places. Do not round any intermediafe calculations.

The probability that two of the three spend less than an hour a night studying is

Answer 1

The number of students being large, the answers observed are almost independent. So it is safe to assume that all three each have 18% chance independently of each other. So we can use binomial probabilities.

Recall that if we have n independent trials with probability of success p then the probability of k successes is C(n,k)p^k(1-p)^n-k.

(a) P(All three) = C(3,3) (0.18)³(0.82)⁰ = 0.005832

Rounded: 0.0058

(b) P(Two of three) = C(3,2) (0.18)²(0.82)¹ = 0.079704

Rounded: 0.0797