# An urn with 5 balls.

I am trying to fully understand the difference between two very similar problems. First I will state the problems. The first I will refer to as problem A and the second I will refer to as problem B.

Problem A: An urn contains 5 balls. Each ball can be either white or black only and the 5 balls the urn contains are equally likely to include 1, 2, 3, 4 or 5 white balls. Gus withdraws 2 balls and they are both white. What are the chances the remaining 3 balls are also white?

Problem B: An urn contains 5 balls each likely to be white or black. Gus withdraws 2 balls and they are both white. What are the chances the remaining 3 balls are also white?

The answer to A requires (in my answer anyway) the use of Bayes theorem. Before the observation that we have drawn two white balls, the probability of obtaining 5 white balls would be ⅕. After the selection of two white balls that probability increases to ½. That is, the observation makes a difference to the outcome.

The answer to B is similar to flipping five coins. After flipping two coins and getting two heads what is the probability of getting 3 further heads? Well that is independent of what has came before and so it is ⅛. I could have said something like “On a table are placed 5 fair coins. Two are flipped and both reveal heads, what is the chances of getting 3 further heads?”. That is, the observation makes no difference to the outcome.

Am I thinking about these problems correctly? I don’t think I fully understand the difference between these problems. On first looking at B I needed a hint that what comes before is irrelevant. Then it all came together.

Is there another way to look at the relationship between these problems? I seem to think that the main difference in these problems is that in A the observation changes the probabilities and in B it does not. By why is that the case? Why aren’t things different after the observation in B? I am autistic, which might be at fault here. I need a different way to think about this.

• Problem A is missing data on the probability of a ball being black or white. Commented Jul 8 at 14:06
• Clarification requested for Problem A : "Each ball can be either white or black only and the 5 balls the urn contains are equally likely to include 1, 2, 3, 4 or 5 balls" - what does this mean? Commented Jul 8 at 15:42

The main difference, as you mentioned, is the dependence. In B, drawing balls doesn't change the probability of the remaining ones at all. The reason is that, since we don't know how many balls of each color we have, the chances of getting a white ball is the same, that's the information we know about B that makes it different from A, which has fixed, known colors.

• Please see the comment that I left following the posted question. How did you determine the specifications for Problem A? Commented Jul 8 at 15:40
• Since A doesn't mention the probability, I'd assume each color has a probability of exactly 0.5. This is obviously not true for all cases, but assuming Gus drew two white balls, we can then find the probability that the remaining balls are all white, and so on. Bayes would be the ideal method. Commented Jul 8 at 19:07

It may be helpful to realize that in problem B, the color of the balls is independent, so learning the color of one cannot help you make inferences about the others.

If we assume independence between the color of individual balls, the total number of white balls must follow a binomial distribution.

In problem A, the number of white balls does NOT follow a binomial distribution and hence the colors have to be dependent.

Specifically, in problem A, there is low probability of choosing two white balls, if the total number of white balls is low. Hence drawing two white balls reduces the probability that there indeed is just a few white balls. But if there are 5 balls, drawing two white balls is totally expected, hence the increased probability.

Additionally, in problem B, the probability that an individual ball is white is already known (given), so the two balls you've drawn provide no extra information about the proportion of white balls. If the probability was unknown, drawing two white balls would make it more likely that the probability of being white is high, so you would consider it more likely to gain three white balls then before.

• Please see the comment that I left following the posted question. How did you determine the specifications for Problem A? Commented Jul 8 at 15:40
• @user2661923 I made a guess that the word "white" is missing as this makes the Bayes theorem calculation after that check out. Commented Jul 8 at 16:03
• You also (apparently) assumed that the probability of $~0~$ white balls was $~0.~$ This seems a reach; you are assuming that the OP correctly interpreted the problem during a portion of their analysis, but not when the OP was stating the problem. Commented Jul 8 at 16:30