Question

An urn contains 6 red balls and 4 green balls. Three balls are chosen randomly from the urn, without replacement.

(a) What is the probability that all three balls are red? (Round your answer to four decimal places.)

(b) Suppose that you win $20 for each red ball drawn and you lose $10 for each green ball drawn. Compute the expected value of your winnings.

Answer 1

(a)

Probability that all three balls are red = Probability that 1st ball is red *  Probability that 2nd ball is red *  Probability that 3rd ball is red = (6/10) * (5/9) * (4/8)

= 0.1667

(b)

Probability that all three balls are red = 0.1667

Total winnings with all three balls are red = $20 * 3 = $60

Probability that all two red balls one green ball = (Number of ways to select two red balls from six *  Number of ways to select one green ball from four) / Number of ways to select three balls from ten

= (⁶C₂ * ⁴C₁ ) / ¹⁰C₃ = (15 * 4) / 120 = 0.5

Total winnings with two red balls and one green red ball = $20 * 2 - $10 * 1 = $30

Probability that all one red ball and two green balls = (Number of ways to select one red ball from six *  Number of ways to select two green balls from four) / Number of ways to select three balls from ten

= (⁶C₁ * ⁴C₂ ) / ¹⁰C₃ = (6 * 6) / 120 = 0.3

Total winnings with two red balls and one green red ball = $20 * 1 - $10 * 2 = $0

Probability that all three balls are green = Probability that 1st ball is green *  Probability that 2nd ball is green *  Probability that 3rd ball is green = (4/10) * (3/9) * (2/8)

= 0.0333

Total winnings with all three balls are green = -$10 * 3 = -$30

Expected value of your winnings = 0.1667 * $60 + 0.5 * $30 + 0.3 * $0 - 0.0333 * $30

= $24.003