# Urn problem with/without replacement

1. Sep 15, 2013

### Thunderbird88

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

Suppose an urn contains 6 white and 9 black numbered balls.

a) 4 balls are collected in random without replacement. What is the probability the first two are white and the last two are black?

b1) 4 balls are collected in random with replacement. What is the conditional probability the first and third balls are black, given that the quartet contains exactly 3 black balls.

b2) The same as b1, except now we collect the balls without replacement.

2. Relevant equations

The conditional probability $$P(A|B) = \dfrac{P(A \cap B)}{P(B)}$$ and the permutations without replacement $$\dfrac{n!}{n1 \cdot n2 \cdot \cdot nk}$$

3. The attempt at a solution

a) I compute $$P(WWBB)=\dfrac{6}{15} \cdot \dfrac{5}{14} \cdot \dfrac{9}{13} \cdot \dfrac{8}{12} = 0.0659$$

b1) Suppose the events
A: the first and third balls are black
B: there are exactly 3 black balls in the quartet

The probability of B is:

$$P(B) = (\dfrac{9}{15})^3$$ , since each time we take out a ball, we replace it, thus the probability is the same.

The probability of $$A \cap B$$ is:

$$P(A \cap B) = (\dfrac{9}{15})^3 \cdot (\dfrac{6}{15})$$

So, the conditional probability is:

$$P(A|B) = \dfrac{P(A \cap B)}{P(B)} = \dfrac{6}{15}$$

b2) Here, the event B has two possibilities, either BBBW or BWBB. However, since we have multiplication, each case has the same probability, so:
$$P(B) = \dfrac{9}{15} \cdot \dfrac{8}{14} \cdot \dfrac{7}{13} \cdot \dfrac{6}{12} = 0.00923$$

The total number of ways that the 15 balls can be arranged is found by:

$$N=\dfrac{15!}{6! \cdot 9!} = 385$$

Since, the possible arrangements we want are BBBW and BWBB, I find:

$$P(A \cap B) = \dfrac{2}{385} = 0.0052$$

So, the conditional probability is:

$$P(A|B) = \dfrac{0.0052}{0.0923} = 0.0056$$

Is my solution correct?

Thank you.

2. Sep 15, 2013

### Ray Vickson

Your answer for (a) is OK, but I, personally, would like more explanation, rather than just seeing some numbers written down---because it is not clear whether you understand the way to do it or are just writing down some more-or-less random numbers. Just a few words of explanation taking ≤ 15 seconds to write would suffice.

Your values for P(B) in (b1) and (b2) are incorrect, so I just stopped checking at that point.

3. Sep 15, 2013

### willem2

B1) It seems you didn't notice that you need to draw 4 balls and 3 of them have to be black. your computation for P(B) is right for getting 3 black balls for 3 draws.

I can't see how you got P(A$\cap$B). As you notice when trying to solve b2, there are only 2 ways to draw the balls that get you both an A and a B event, and those are BWBB and BBBW. The probabilites of those shouldn't be hard to find if you did solve problem a

B2) What are your A and B events here? It seems logical to use the same events you used in B1.
Your computation for P(B) looks like a computation for P(A$\cap$B), but you forget that there are 2 cases, and you have an extra 0 after the decimal point.

Your computation for P(A$\cap$B) is incomprehensible.
This should be the same computation as B1, except for the fact that the probabilities for each draw change because of the removed balls.

4. Sep 15, 2013

### HallsofIvy

Staff Emeritus
I personally wouldn't use formulas (I can never remember them!). Originally, the urn contains 6 white and 9 black balls, total of 15. The probability the first is white is 6/15= 2/5. Now there are 14 balls, 5 white and 9 black. The probability a white ball is drawn is 5/14. Now there are 13 balls, 4 white and 9 black. The probability a black ball is drawn is 9/13. Finally, there are 12 balls, 4 white and 8 black. The probability a black ball is drawn 8/12= 2/3.

The probability "the first two are white and the last two are black" is the product (2/5)(5/14)(9/13)(2/3).

Since this is done "with replacement", the probability any specific ball drawn is black is 9/15= 3/5. Given that there are 3 black and one white ball, there are 4 possible places the white ball could be, all "equally likely", and the white ball is in the first position in one and the third position in another so the other two possibilities have a black ball in both positions. The probability is 2/4= 1/2.

Since we are given that we drew three black and one white ball, the fact that this is "without replacement" is irrelevant. There are $_4C_3$= 4 ways to order "three black and on white ball" and there are black balls in the first and third places in two of them. The probability is still 2/4= 1/2.