Question 1193829: 12 balls — 4 red, 4 white, and 4 blue.
If three balls are drawn at random out of this box one at a time without replacement, what is the probability of obtaining
3 white balls?
Answer: 1/55
4/12 * 3/12 * 2/10
3 balls of the same color?
Answer: 3/55
p{(w,w,w),(R,R,R),(B,B,B)}
1/55 + 1/55 + 1/55
Are my answers correct? Thanks
Found 2 solutions by Edwin McCravy, math_tutor2020:
Answer by Edwin McCravy(20055)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Question 1: What is the probability of getting 3 white balls?
If we select without replacement, then,
4/12 is the probability of getting a white ball
3/11 is the probability of getting a second white ball
2/10 is the probability of getting a third white ball
Each time we count down by 1 for the numerator and denominator separately.
Multiply out those fractions:
(4/12)*(3/11)*(2/10)
(4*3*2)/(12*11*10)
(4*3*2)/(4*3*11*2*5)
1/(11*5)
1/55
You have the correct answer. Nice job.
Here's another way to get the answer.
There are n = 4 white balls and we select r = 3 of them.
Using the nCr combination formula leads to 4C3 = 4 ways to do this. In other words, there are 4 ways to not select a particularly unlucky white ball.
Now let's count the number of ways to select any 3 balls regardless of color.
We have n = 12 of them and we select r = 3
12 C 3 = 220 ways to pick any 3 balls
Then notice how 4/220 = (1*4)/(55*4) = 1/55
Here's some scratch work to show how to compute 12 C 3
n C r = (n!)/(r!(n-r)!)
12 C 3 = (12!)/(3!*(12-3)!)
12 C 3 = (12!)/(3!*9!)
12 C 3 = (12*11*10*9!)/(3!*9!)
12 C 3 = (12*11*10)/(3!)
12 C 3 = (12*11*10)/(3*2*1)
12 C 3 = (1320)/(6)
12 C 3 = 220
Use this template to compute 4C3 as well if necessary.
As you can tell or already guess, order does not matter when selecting the balls which is why the nCr formula is used (rather than the nPr one).
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Question 2: What is the probability of getting 3 balls of the same color?
You are correct in thinking that 1/55 is the probability of getting all red, and also all blue. This works because we have 4 of each color.
This leads to the correct answer of 3/55. Once again, nice work.
Side note: I appreciate you showing your work and thought process.
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