# Probability That They Are One Red, One White And One Blue Balls

1. Jul 10, 2012

### chikis

1. The problem statement, all variables and given/known data

Three balls are drawn one after the other with replacement, from a bag containing 5 red, 9 white and 4 blue identical balls. What is the probability that they are one red, one white and one blue?

2. Relevant equations

3. The attempt at a solution

The question demands that I evaluate the probability that they are one red, one white and one blue balls in the three draws held.
They did not give the order which the draws will follow, I therefore assume it could be in order.
Here is my working:
Total number of balls = 18
P(r) = 5/18
P(w) = 9/18 = 1/2
P(b) = 4/18 = 2/9
The probability that they are one red, one white and one blue balls = P(1st-r,2nd-w,3rd-b) Or P(1st-w,2nd-b,3rd-r) Or P(1st-b, 2nd-r,3rd-w)
(5/18*1/2*2/9) + (1/2*2/9*5/18) + (2/9*5/18*1/2)
= 10 + 10 + 10/324
= 30/324
= 5/54
But the answer I got did not match with the answer they provided for the question. Is my working wrong or did I not follow the right principle?

2. Jul 10, 2012

### Villyer

You mentioned three drawing orders, r.w.b, w.b.r, b.r.w.

3. Jul 10, 2012

### LCKurtz

You might find it easier to not treat this as though order counts. Have you studied the hypergeometric distribution? Look at

http://en.wikipedia.org/wiki/Hypergeometric_distribution

for a discussion and you will see how to use it for your problem.

4. Jul 11, 2012

### chikis

But the questions says, "Three balls are drawn one after the other with replacement, from a bag", so assume they three draws because they three balls. Don't you think the same?

5. Jul 11, 2012

### LCKurtz

Sorry. I mis-read it as without replacement.

6. Jul 11, 2012

### chikis

Even if it is without replacement, does that make a change to the already laid down principle?

7. Jul 11, 2012

### HallsofIvy

Staff Emeritus
With replacement, there are, at each draw, 18 balls, 5 red, 9 white, and 4 blue. The probability of drawing a red ball is 5/18, a white ball is 9/18, and a blue ball is 4/18. Of course, because this is with replacement these are independent. If A and B are independent events then P(A and B)= P(A)P(B).

However, there are 3!= 6 "orders" in which you can draw the three balls, not 3:
RWB
RBW
WRB
WBR
BRW
BWR

Instead of calculating each one separately and adding note that the probabilities of each are all the same- (5/18)(9/18)(4/18)- and multiply by 6.

8. Jul 11, 2012

### chikis

I think it is important at this juncture, that I inform you that your system of working is quite different from the working and answers provided by the people who set the question.
Here is the list of possible answers based on how you do the workings. A. 5/102 B. 5/136 C. 5/162 D. 5/204 E. 5/243.
It is unforturnate to note that your working did not fall into any of these answers.
Here is their own working:
Total number of balls = 18
P(1R, 1W, 1B) = 5/18 * 1/2 * 2/9 = 5/162 = option C.
What do you have say about it?

9. Jul 11, 2012

### chikis

I think it is important at this juncture, that I inform you that your system of working is quite different from the working and answers provided by the people who set the question.
Here is the list of possible answers based on how you do the workings. A. 5/102 B. 5/136 C. 5/162 D. 5/204 E. 5/243.
It is unforturnate to note that your working did not fall into any of these answers.
Here is their own working:
Total number of balls = 18
P(1R, 1W, 1B) = 5/18 * 1/2 * 2/9 = 5/162 = option C.
What do you have say about it?

10. Jul 13, 2012

### LCKurtz

That calculation looks correct since the three events are independent.

11. Jul 14, 2012

### HallsofIvy

Staff Emeritus
That is correct if the original question was "one red, one white and one blue ball" in that order. That was not my interpretation since the words "in that order" were not in the original problem. In fact you said, in your original post, "They did not give the order which the draws will follow".

In any case, it is not "quite different"- it only drops the factor of "3!= 6", the number of possible orders.