Question

An urn contains 4 red, 5 blue, 2 yellow, and 9 green balls. A group of 5 balls is selected at random without replacement. What is the probability that the sample contains:

a. atleast one of each color?

b. at most 3 green balls?

Answer 1

(a)

Red = 4

Blue = 5

Yellow = 2

Green = 9

Total balls = 20

Number of balls selected = 5

Number of ways of selecting 5 balls from 20 balls = 20C5 = 15504

P(Red = 2, Blue =1, Yellow= 1, Green = 1)= 4C2 X 5C1 X 2C1 X 9C1/20C5 = 6 X 5 X 2 X 9/15504 = 0.0348

P(Red = 1, Blue =2, Yellow= 1, Green = 1)= 4C1 X 5C2 X 2C1 X 9C1/20C5 = 4 X 10 X 2 X 9/15504 = 0.0464

P(Red = 1, Blue =1, Yellow= 2, Green = 1)= 4C1 X 5C1 X 2C2 X 9C1/20C5 = 4 X 5 X 1 X 9/15504 = 0.0116

P(Red = 1, Blue =1, Yellow= 1, Green = 2)= 4C1 X 5C1 X 2C1 X 9C2/20C5 = 4 X 5 X 2 X 36/15504 = 0.1187

So,

P(at least one of each color) = 0.2115

So,

Answer is:

0.2115

(b)

P(Green = 0, Others = 5) = 11C5/20C5 = 462/15504 = 0.0298

P(Green = 1, Others = 4) = 9C1 X 11C4/20C5 = 9 X 330/15504 = 0.1916

P(Green = 2, Others = 3) = 9C2 X 11C3/20C5 = 36 X 165/15504 = 0.3831

P(Green = 3, Others = 2) = 9C3 X 11C2/20C5 = 84 X 55/15504 = 0.2980

So,

P(at most 3 green balls) = 0.9025

So,

Answer is:

0.9025