# Dragging colored balls off box in succession

1. Apr 15, 2015

### ChrisVer

Suppose you have a box with 10 balls, the 4 are red, the 5 are blue and the 1 is white.
The probability of dragging each ball is $0.4~,~0.5~,~0.1$ respectively.

Now if I ask what is the probability of getting 3 balls in the order: red -> blue -> white, what is the right answer if balls are replaced/not replaced?

I have some problem in understanding the logic behind my approach to answer this...
first I thought: $P_{tot} = P(R) P(B) P(W)$ This gives the probability of dragging a red, a blue and a white ball... It doesn't seem right for ordered draggings since it also counts the dragging of : blue->red->white.
If the balls are replaced, then the dragging doesn't affect the probabilities, and so the correct probability of the successive drawing would be:
$P(RBW)= \frac{P(R) P(B) P(W)}{3 !}= \frac{0.02}{6} = \frac{1}{300}$

Since the written $P_{tot} = P(RBW)+ P(RWB)+ P(BRW) + P(BWR) + P(WBR) + P(WRB)$ (permutations of 3).

Then I thought of using the conditional probability...I define two events: $A_1,A_2$. The $A_1$ is the event that I drag a blue ball given that I had dragged a red ball before: $A_1 = (B|R)$ and the probability is:
$P(A_1)= P(B|R)= \frac{P(B \cap R)}{P(R)}= \frac{P(B) P(R)}{P(R)} = P(B)$

Then $A_2$ is dragging a white ball, given that $A_1$ is true:
$P(A_2)= P(W|A_1)= \frac{P (W \cap A_1)}{P(A_1)} = P(W)$
and so the $P(RBW)=P(W)=0.1 \ne \frac{1}{300}$
I used that $P(x \cap y) = P(x) P(y)$ since they are statistically independent events.

So what is wrong in either of these approaches? They obviously don't give the same result...However, logically they make sense to me

Last edited: Apr 15, 2015
2. Apr 15, 2015

### HallsofIvy

You need to be clearer on the precise situation. You say you have "10 balls, the 4 are red, the 5 are blue and the 1 is white." If you pick balls at random then the probability the first ball you pick is red is 4/10= 0.4, blue is 5/10= 0.5 and white is 1/10= 0.1. Now, are you picking additional balls "with replacement" or "without replacement"?

If this is "with replacement" then you put the ball back so that the probabilities remain 0.4, 0.5, and 0.1 But if it is "without replacement" those will change.

So "with replacement" "Red, Blue, White", in that order, is (0.4)(0.5)(0.1). If this is "with replacement" then the probability the first ball is red is 4/10= 0.4. Then there are nine balls, five of them blue so the probability the second ball will be blue is 5/9. Then there are 8 balls, 1 of which is white, so the probability the last ball is white is 1/8. The probability of "Red, Blue, White", in that order, without replacement, is (4/10)(5/9)I1/8).

3. Apr 16, 2015

### ChrisVer

I just gave the idea for the replacing the balls case and the probabilities remaining the same...
The thing is that I don't understand why the two approaches (multiplying the probabilities and using the conditions to ensure the ordering), which seem both plausible for me, give different results... so one must be wrong (judging from your answer, the conditional probabilities are wrong) and I don't see why...

Also shouldn't there be the 3! in your (0.4)(0.5)(0.1) ? (for the reasoning I used in $P_{tot}$ )

4. Apr 16, 2015

### HallsofIvy

The two approaches give different answers because they are different situations! Look at these problems: you have 10 balls, 8 white and 2 black. With replacement what is the probability of getting three consecutive black balls? The answer is, of course, $(2/10)^3= 0.008$. What is the probability without replacement? It should be immediately obvious that, since there are only two black balls, the probability of getting three is 0! If you do not see the difference then I suspect it is a language problem.

Last edited by a moderator: Apr 16, 2015
5. Apr 16, 2015

### Orodruin

Staff Emeritus
No, this gives the probability of one particular order. The probability of the first ball you take being red is P(R), you then move on to the second and the probability of drawing a blue is P(B), then to the last and the probability of drawing white is P(W). Assuming these are independent events, the full probability of drawing the balls in that order is P(R)P(B)P(W). This is essentially what you did in the second approach. Therefore, you should not divide by 3!.

However, if you do not care about the order, you could have done this in 3! different ways and the probability would be 3! larger, i.e., 3! P(R)P(B)P(W).

The take home message is that P(R)P(B)P(W) is the probability of drawing one red, one blue, and one white in a given order. (Assuming replacement. Otherwise the draws are not independent.)

Edit:
No, there shouldn't be, because the reasoning is wrong.

6. Apr 16, 2015

### ChrisVer

I am talking about the one with replacement...(so no need to mention the without replacement)
So we have that the probability of drawing a r+b+w is $P(R) P(B) P(W)= 0.02$

However, if I use conditional probabilities:
draw white given that blue was drawn (and blue given that red was drawn), gives:
$P(W|(B|R))=P(W)=0.1$ (see post 1)

That's my problem. The only thing that can make this equation not be correct is that the drawings are not statistically independent - but that doesn't make sense in the given scenario (of balls replacement)

The language of both seems equivalent: draw first red, second blue , third white = draw final white given that blue was drawn given that red was drawn.

PS. I got the 3! reasoning...

Last edited: Apr 16, 2015
7. Apr 16, 2015

### Orodruin

Staff Emeritus
This is the probability of getting a white given that you have already gotten R and B in that order. This is not equal to the probability of drawing R, B, W in that order. Note that P(AB) = P(A|B) P(B). If A is the event drawing W in the last ball and C is "drawing RB with the first two", we would obtain:
P(RBW) = P(AC) = P(A|C) P(C) = P(A)P(C)
since A and C are independent. That is, you have to also account for the probability that you may not arrive at the situation where you have drawn RB with the first two.

Edit: Seen in a different way. Let A be the probability of drawing W after drawing RB. We then generally have:
P(A) = P(A|C)P(C) + P(A|not C) P(not C)
Now P(A|C) is just the probability of drawing white, i.e., P(W). P(A|not C) is equal to zero because it is not possible to draw W after drawing RB if you have not drawn RB.

8. Apr 16, 2015

### HallsofIvy

Yes, P("r+ b+ w") is P(R)P(B)P(W)= 0.4(0.5)(0.1)= 0.02

Yes, that is correct. why is that a problem? P(W|(B|R))= P(W) because the drawings are independent.

I really don't understand why you have a problem with this. Yes, both say "draw first red, second blue, third white" And both can be done as "P(White given (Blue give Red). The difference in the two problems is that "P(Blue given Red)" is different for the two problems and then P(White given (Blue given Red)) is different for the two problems..
In the first problem, with replacement P(Blue given Red) is equal to P(Blue)= 5/10= 1/2. while without replacement, P(Blue given Red) is 5/9 because, having drawn a red ball the first time, there are now 3 red balls, 5 blue balls, and one white ball left.

9. Apr 16, 2015

### ChrisVer

The problem was that $0.1 \ne 0.02$.

But I think it is answered now, I'm trying to look through it.

10. Apr 16, 2015

### HallsofIvy

And thank goodness they are not! 0.1 is the probability of drawing a white ball. 0.02 is the probability of drawing a red, a blue, and a white ball in that order. Those are completely different "outcomes". (Notice that in order to get "red, blue, and white, in that order" we certainly must get "white" which is why "probabllity of red, blue, and white, in that order"= 0.02 is less than "probability of white"= 0.1.)