Question

An urn contains 4 red balls and 6 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 $50 for each red ball drawn and you lose $25 for each green ball drawn. Compute the expected value of your winnings.

Answer 1

Number of ways to select r items from n, nCr = n!/(r! x (n-r)!)

Number of red balls = 4

Number of green balls = 6

Total number of balls = 10

a) P(all three balls are red) = Number of ways to select 3 red balls from 4 / Number of ways to select any 3 balls from 10

= 4C3 / 10C3

= 4/120

= 1/30

= 0.03333

a) P(0 red balls) = 6C3/10C3

= 20/120

= 1/6

P(1 red ball) = 4C1 x 6C2 / 10C3

= 4 x 15 / 120

= 1/2

P(2 red balls) = 4C2 x 6C1 / 10C3

= 3/10

P(3 red balls) = 1/30

Number of red balls, X	0	1	2	3
Winning amount	3x-25 = $-75	50 - 2x-25 = $0	2x50 - 25 = $75	3x50 = $150
P(X)	1/6	1/2	3/10	1/30

Expected value of your winnings = -75x1/6 + 0x1/2 + 75x3/10 + 150x1/30

= $15