SOLUTION: A bag contains three black balls, four white balls and five red balls. Three balls are removed without replacement. What is the probability of obtaining 1. One of each colour 2.

Question 1156979: A bag contains three black balls, four white balls and five red balls. Three balls are removed without replacement. What is the probability of obtaining
1. One of each colour
2. At least two red balls
3. At most two white balls
(show the posibilities)
Answer by VFBundy(438) (Show Source): You can put this solution on YOUR website!
In total, there are 12 balls. Since you are choosing three balls without replacement, there are THIS many ways in which you can choose three balls:

12C3 = = 220

You will use this in each of the problems below.

1. One of each color

The number of ways you can choose one black ball, one white ball, and one red ball:

3C1 * 4C1 * 5C1 = = 3*4*5 = 60

The probability of choosing one of each color = 60/220 = 3/11

2. At least two red balls

The number of ways you can choose three red balls OR two red balls and one black ball OR two red balls and one white ball:

5C3 + (5C2 * 3C1) + (5C2 * 4C1) = + + = 10 + (10*3) + (10*4) = 10 + 30 + 40 = 80

The probability of choosing at least two red balls = 80/220 = 4/11

3. At most two white balls

The easiest way to do this would be to figure out the number of ways you can choose three white balls, then subtract this amount from 220.

The number of ways you can choose three white balls:

4C3 = = 4

Since choosing "at most two white balls" would be every possibility EXCEPT choosing three white balls, you simply subtract the number of ways you can choose three white balls (4) from the total number of ways you can choose any three balls (220). This means there are 216 ways of choosing "at most two white balls."

The probability of choosing at most two white balls = 216/220 = 54/55
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