Ask Question
20 November, 08:04

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

+4
Answers (1)
  1. 20 November, 08:15
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Brian, a landscape architect, submitted a bid on each of three home landscaping projects. He estimates that the probabilities of winning ...” in 📗 Mathematics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers