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

[chikis]

Homework Statement

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?

Homework Equations

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?

 

[Villyer]

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

What about others?

 

[LCKurtz]

Homework Statement

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?

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.

 

[chikis]

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

What about others?

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?

 

[LCKurtz]

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?

Sorry. I mis-read it as without replacement.

 

[chikis]

Sorry. I mis-read it as without replacement.

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

 

[HallsofIvy]

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.

 

[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?

 

[chikis]

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

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?

 

[LCKurtz]

rs.
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?

That calculation looks correct since the three events are independent.

 

[HallsofIvy]

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?

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.