# If 3 red balls and 3 blue balls are numbered 1-6, is it possible to calculate the probability of randomly selecting an even numbered red ball?

Consider the following question from Manhattan Prep's GRE kit. Here's an image with the question and answer given by Manhattan prep. I'll write the question below as well:

Six billiard balls, numbered 1 to 6, are placed in a box. Three of them are colored red, and three of them are colored blue. One ball is randomly drawn from the box. What's the probability that the ball drawn will be an even numbered red ball?

According to manhattanprep, the probability cannot be computed since we don't have any information about the distribution of colors among the numbered balls. In my opinion, the probability can be calculated by identifying the possible types of color assignments that the balls can have, and considering every possible assignment, calculate the conditional probability of getting a red even ball, and sum that up for all possible assignments. I've written my approach in detail here.

Lets say there are 10 balls, numbered 1 to 10, in jar A. Person P1 picks 2 balls from jar A and puts them in jar B. Person P2 picks 1 ball from jar B. What's the probability that the ball chosen by P2 is ball no. 5?

Will the probability be different if person P1 knows that the 2 balls he picked didn't have the ball no.5 at all?

According to me, it matters as to who is computing the probability, and what information he has while doing that. If P1 already knows there wasn't ball no. 5 at all, according to him P2 has zero chances of picking up ball no. 5. But if P2 doesn't have any idea of what P1 picked, he is going to assume that he has some non zero chance of picking ball no. 5, and he can compute that chance.

Isn't this case similar to the above question? where P2 computes his chances of getting ball no. 5 by calculating the possible distribution of numbers among the 2 balls that are there in jar B and calculating his chances from there on.

• Which numbers correspond to red balls is not random, however. But you're treating it as if it were.
– user296602
Sep 14, 2017 at 15:47
• What is the probability that Winston Churchill was a Scorpio? Sep 14, 2017 at 16:20

In situation one, your detailed approach is imposing an additional level of randomization onto the experiment -- namely, that the assignment of colors to numbers is also performed randomly, and specifically that all ${6\choose 3}=20$ ways to select 3 balls to 'receive' the red color are equally likely. If this assumption is made, then your analysis is correct. But this assumption wasn't made explicit in the problem statement. The furthest you can get in solving the problem is to compute the probability conditional on each of your four scenarios $A$, $B$, $C$, $D$; it's not stated in the problem what the probabilities $P(A)$, $P(B)$, $P(C)$, $P(D)$ are, so you can't proceed to the final unconditional probability.

As for your second situation, yes, the probability being computed depends on the information you have (or are allowed to assume). Knowing that P1 didn't pick ball no.5 corresponds to a conditional probability: $P(\text{P2 picks ball 5}\mid\text{ball 5 not in jar$B$})$. OTOH, without this knowledge you are expected to compute the unconditional probability $P(\text{P2 picks ball 5})$. Note, however, that in contrast to your first situation, the unconditional probability can be computed unambiguously, without assuming additional randomization, because the description of the experiment being conducted allows you to compute $P(\text{ball 5 in jar$B$})$ and $P(\text{ball 5 not in jar$B$})$.

"According to manhattanprep, the probability cannot be computed since we don't have any information about the distribution of colors among the numbered balls."

Nitpicky. Pendantic. And a smart-ass.

But technically correct. We live in a universe where the balls were painted at the factory and can not change.

" In my opinion, the probability can be calculated by identifying the possible types of color assignments that the balls can have, and considering every possible assignment"

And manhattenprep is saying there is only one possible type of color assignment; the one that happened. All others did not happen and are not possible.

"Isn't this case similar to the above question?"

If we were told that the factory used a randomly equal distribution method of painting the balls, yes, which is what the GRE test was clearly assuming.

In saying you are know there is no five, you are replacing an arbitrary sample with a specific sample. It's the difference of asking what is the probability the an arbitrary person is a Scorpio and asking what is the probability Winston Churchill is a Scorpio.

It's a quibble, but it is a valid quibble.

• The counterpoint is that in the Bayesian interpretation of probability, it is absolutely valid to say that since we don't know anything about how colors were assigned to numbers, to us they are equally likely. Similarly, since I know nothing about Winston Churchill's birthday, to me he is a Scorpio with probability $\frac1{12}$. If you've looked it up, then to you the probability is either $0$ or $1$ (or, technically, $\frac1{12}\epsilon$ or $1-\frac{11}{12}\epsilon$ where $\epsilon$ is the hopefully-tiny chance your source was wrong). Sep 14, 2017 at 16:24