urn with two colors of balls; probability of selecting specific color

10 balls in an urn, 6 Black and 4 White. Three are removed, color not noted. What is probability that a white ball will be chosen next?

The answer is 2/5, so my reasoning below must be faulty.

After the initial three balls are removed, there will be 4 possible configurations:

A: BBB___  WWWW
B: BBBB__  _WWW
C: BBBBB_  __WW
D: BBBBBB  ___W

P(w|A) = 4/7
P(w|B) = 3/7
P(w|C) = 2/7
P(w|D) = 1/7


Answer should be P(w|A)*P(A) + P(w|B)*P(B) + P(w|C)*P(C) + P(w|D)*P(D)

P(A): remove 1st black ball (p=6/10); remove second black ball (p=6/10 * 5/9); remove third black ball (p = 6/10 * 5/9 * 4/8)= 120/720

P(B): remove 2 black and one white; also 120/720

P(C): 6/10 * 4/9 * 3/8 = 72/720

P(D): 4/10 * 3/9 * 2/8 = 24/120

Doing the math gives me 0.239.

You calculated $P(A)$, $P(B)$, $P(C)$ and $P(D)$ incorrectly. A can happen in $\binom{6}{3} = 20$ ways, B in $\binom{6}{2} * \binom{4}{1} = 60$ ways, C can happen in $\binom{6}{1} * \binom{4}{2} = 36$ ways, D can happen in $4$ ways. To check, there are $20+60+36+4=120$ total ways of removing $3$ balls at random, which is $\binom{10}{3}$.

The answer is then $\frac{4}{7} * \frac{20}{120} + \frac{3}{7} * \frac{60}{120} + \frac{2}{7} * \frac{36}{120} + \frac{1}{7} * \frac{4}{120} = \frac{2}{5}$

You don't actually have to do any calculating. Since no information was added by removing the balls, the probability can't change. No matter how many balls we remove, the probability that the next ball we choose will always be 2/5.

• Is that true (prob that white will be chosen next= 2/5) even if 9 balls were first removed? That is not obvious at first sight to me. Commented Sep 4, 2017 at 12:36
• Think of that this way: You have ten balls. You take any one of them, and put the other nine somewhere else, thereby dividing them into 9 and 1. Now it should not make a difference whether you first draw one and then look at the other nine, or whether you first draw nine, disregard them, and then look at the color of the last one. Both of these actions are a separation of 10 into 9 and 1, and we are looking at the probability that the 1 ball is white. Commented Sep 4, 2017 at 12:39
• Drawing 9 is the same as choosing 1. So, if you choose 1 ball, what is the chance that it is white? Commented Sep 4, 2017 at 12:47
• Perhaps if you write the answer as $\frac{4}{10} = \frac{2}{5}$, the rationale will be clearer. If the $10$ balls are arranged in a row, the probability that the $k^{\text{th}}$ ball is white is $\frac{4}{10}$ regardless of what $k$ is. This is a property that sampling without replacement shares with sampling with replacement; a fact that puzzles most beginners in probability. Commented Sep 4, 2017 at 15:26

Think of it this way:

Take the urn with 10 balls as it is. Draw three, put them to the side. These are the removed ones. Now draw a fourth one.

It should have a probability of 40% to be white, because each ball has a 40% probability to be white and we do not look at the colors of the balls removed earlier.

You could have that the three balls that were removed initially all were white. This would then greatly increase your chance of getting a black one with the fourth draw. However, they could also have all been black, thereby increasing your chance of getting a white one with the fourth draw. It can be proven that these will cancel out and the probability stays unchanged, if you do the math correctly.

As a matter of fact, since the three balls are selected randomly, would you intuitively think that there should be a change in expected color of the fourth one after you took out three balls without looking at their colors?

• Thanks for the intuitive 'vision' approach; I would like to be able to see the big picture at a glance. Re your last sentence: it seems that the 3 prior draws could change or affect the expected color of the fourth one. Commented Sep 4, 2017 at 12:31
• yes, but not if we don't look at the colors of the three balls. I shall make that more clear. Commented Sep 4, 2017 at 12:36