Brian, a landscape architect, submitted a bid on each of three home landscaping projects. He estimates that the probabilities of winning the bid on Project A, Project B, and Project C are 0.8, 0.5, and 0.2, respectively. Assume that the probability of winning a bid on one of the three projects is independent of winning or losing the bids on the other two projects. Find the probability that Brian will experience the following. (a) Win all three of the bids (b) Win exactly two of the bids

a: 0.08, or 8% chance he wins all 3

b: 0.42, or 42% chance he wins 2

Step-by-step explanation:

P (win A) = 0.8

P (lose A) = 0.2

P (win B) = 0.5

P (lose B) = 0.5

P (win C) = 0.2

P (lose C) = 0.8

The situations are independent, so we multiply probabilities together.

To win all 3: P (win A) * P (win B) * P (win C) = 0.8*0.5*.02 = 0.08

To win 2 of the 3 there are 3 ways to do this. We add up the probabilities of the 3 situations ...

P (win A) * P (win B) * (lose C) = 0.8*0.5*0.8 = 0.32

P (win A) * P (lose B) * P (win C) = 0.8*0.5*0.2 = 0.08

P (lose A) * P (win B) * P (win C) = 0.2*0.5*.02 = 0.02

0.32 + 0.08 + 0.02 = 0.42