# A weird probability question

This is the problem in question:

You have two identical bowls: the first one contains 3 white balls and 4 black balls, and the second one contains 4 white balls and 5 black balls. If you choose randomly a ball from the two bowls, what is the probability it is white?

Let's define our events as such:

A1 = choosing a ball from the first bowl

A2 = choosing a ball from the second bowl

B = choosing a white ball

One approach would be using the theorem of total probability:

$$\text{We know that }P(B|A_1) = \frac34\text{ and }P(B|A_2) = \frac45\text{, and that:}$$

$$P(A_1) = P(A_2) = \frac12\text{, because the bowls are identical}$$

$$P(B) = P(B |A_1)\times P(A_1) + P(B|A_2)\times P(A_2) = \frac37 \times \frac12 + \frac49 \times \frac12 = 55/126$$

The second approach would be simplifying the problem:

Because the two bowls are identical, we could just say we don't even choose between two bowls, but just between the set of all balls.

Then we could calculate the probability directly:

$$P(B) = \frac {\text{number of white balls}} {\text{total number of balls}} = \frac7 {16}$$

Now, which one is correct? (and why?)

Both solutions seem reasonable, and they have approximately the same value

$$\frac{55}{126}\approx0.4365$$

$$\frac{7}{16}\approx0.4375$$

However, mathematically speaking, they are different results. Which one is correct?

• What is meant by "choose randomly a ball from the two bowls"? Feb 25 at 23:37

The difference in probabilities arises because the two sampling methods are different.

To understand why, consider an extreme case where one bowl has a single white ball, and the second bowl has $$100$$ black and $$99$$ white balls.

If you use the first sampling method--i.e., choose one of the two bowls equally at random, then choose a ball uniformly at random from that bowl, then it's immediately obvious that you have at least a $$1/2$$ probability of drawing a white ball, because if you choose the bowl with the single white ball, you are guaranteed to draw a white ball. And if you pick the second bowl, you have a $$99/199 \approx 1/2$$ chance of drawing a white ball, so the total probability is roughly $$3/4$$.

But if you use the second sampling method, there are equal numbers of white and black balls and the probability of drawing a white ball is exactly $$1/2$$.

So what is going on here is that the hierarchical sampling approach (choose a bowl, then choose a ball in that bowl) does not give you the same chance of drawing any particular ball, so long as the number of balls in each bowl are different. Such a sampling method, applied to the extreme example, would pick the singular white ball half of the time, whereas in the second sampling method, it would be chosen on average only $$1/200$$ of the time.

So the question of "which probability is correct" depends on what kind of sampling procedure you use. This is why, in the analysis of experimental data, it is important to consider how sampling takes place.

• For an even more extreme example, make all 199 balls in the second bowl black. Feb 25 at 18:13

It depends on what you call “pick a ball at random”.

If you choose any ball at random, it doesn’t matter in which bowl it was. The probability is equal to the amount of all while balls divided by the amount of all balls, so $$7/16$$.

Note that bowls are not equiprobable here. You will more likely pick a ball from a bigger bowl.

If you however pick a bowl at random with probability $$1/2$$ then your first calculation is right.

The wording of the problem is ambiguous. If we interpret it to mean you will be selecting the bowl first, and then a ball from that bowl, then your first approach is correct. The second one is incorrect. Let me explain.

Because the two bowls are identical, we could just say we don't even choose between two bowls, but just between the set of all balls.

Are the two bowls identical? If we are talking about the probability of selecting one of he two bowls to draw from, then yes. But they are not identical in the sense that you can consider their contents together. The proportion of white balls in the first bowl is smaller than that in the second bowl. so no, they are clearly not identical.

• Great answer! I don't know how I didn't think about that. Also an edge case would be if we had 99 white balls in the first one and 1 black ball in the second. Because choosing either bowl are equiprobable, that means it's a 50% chance of a white ball. But if we use my second approach, 99/100 = 99% chance of white, obviously wrong Feb 24 at 20:49
• Thank you, glad to be of help. And yeah, that example really drives home the point. And this is also an example of why it's important to be precise in wording, especially in combinatorics and probability. As others have pointed out in their answers, if it's not compulsory to choose a bowl and then a ball from that bowl, the answer would be different. Maybe we empty out the contents of the two bowls and then pick a ball from the whole collection. Or maybe the balls are numbered and we pick a random number without even approaching he bowls. Feb 24 at 21:16

I would agree the wording is somewhat ambiguous. However, if we assume a crazy person is not writing this problem, then I think the intended meaning is that the two bowls are identical in every way except for having different amounts of different colored balls. Furthermore, I doubt the existence of two bowls is irrelevant to the problem. Otherwise, why mention it? To be safe, you could provide your teacher with two answers, one for each possible interpretation, and be sure to explain why each answer is relevant to the corresponding interpretation.

I am partial to the following interpretation, and here is how I would approach it from first principles:

Let $$W$$ be the event in which a white ball is chosen, and let $$B_1$$ and $$B_2$$ be events in which a ball is selected from bowls $$1$$ and $$2$$, respectively. Furthermore, let $$W \cap B_1$$ be the joint event of choosing a white ball from bowl $$1$$, and let $$W \cap B_2$$ be the joint event of choosing a white ball from bowl $$2$$.

First, recognize the probability of choosing a white ball is the probability of either choosing a white ball from bowl $$1$$ or choosing a white ball from bowl $$2$$. Hence, we can express the probability of choosing a white ball as the probability of the union of two joint events

$$P(W) = P((W \cap B_1) \cup (W \cap B_2) )$$

Since the two events $$W \cap B_1$$ and $$W \cap B_2$$ are disjoint, we have

$$= P(W \cap B_1) + P(W \cap B_2)$$

and by the mulitplication rule we have

$$= P(B_1)P(W|B_1) + P(B_2)P(W|B_2)$$

Now, if the choice of bowl is random, then we have $$P(B_1) = P(B_2) = \frac{1}{2}$$. If the choice of ball within a given bowl is random, then we have $$P(W|B_1) = \frac{3}{7}$$ and $$P(W|B_2) = \frac{4}{9}$$. Thus, by substitution we have

$$= \frac{1}{2} \cdot \frac{3}{7} + \frac{1}{2} \cdot \frac{4}{9} = \frac{3}{14} + \frac{4}{18} = \frac{27+28}{126} = \frac{55}{126} \approx 0.4365079365$$

The main problem is just that, if we restrict to reality, and thus perhaps intuition, you cannot truly pick balls at random when they are in two bowls. You have to pick the bowls first, which is the first method.

The second "method" is equivalent to assuming the probability of a bowl being picked is proportional to the number of balls in them. This additional (equivalent) assumption is what makes the result different. And that cannot make sense in reality unless you destroy the bowls first.

• There is a similarity between this problem and the issue in national elections in a country that is divided in regions that are unequally populated.
– Stef
Feb 25 at 21:48

The probabilities $$P(B|A_1)$$ and $$P(B|A_2)$$ are initially misstated, as $$\frac{3}{4}$$ and $$\frac{4}{5}$$ respectively, but should be $$\frac{3}{7}$$ and $$\frac{4}{9}$$ as correctly used in the subsequent calculations.

Your confusion stems from the fact that the problem is not properly stated as a probability computation. For that, you need to have a probability space $$\Omega$$, and a probability measure on that space. The space must be such that every point in $$\Omega$$ completely determines the final outcome of the experiment, so that the latter can be viewed as realised by a single drawing of a point from$$~\Omega$$, and in the case where $$\Omega$$ is at most countable (as it can be here, in fact there are only finitely many outcomes) the probability measure can be given by simple assigning a probability (from the interval $$[0,1]$$) to each point of$$~\Omega$$ so that the sum of the probabilities is$$~1$$. In simple problems $$\Omega$$ can be taken to be the set of all possible final outcomes, and if $$\Omega$$ is specified and finite, there is a probability measure that is easy to describe which is the uniform measure, giving the same probability $$\frac1n$$ to each point of $$\Omega$$ where $$n$$ is their number. But this is not the only measure possible, and it can only be used when $$\Omega$$ is clear. Just saying something is chosen "at random" does not imply that all choices are equally probable.

So when you say a ball is chosen at random, that might suggest taking $$\Omega$$ to be the set of $$16$$ balls with a uniform probability. In that case the event of choosing a ball from the first bowl has probability $$7/16$$, the one of choosing one from the second bowl has probability $$9/16$$ (so it is more likely in this scenario), the event of choosing a white ball has probability $$(3+4)/16$$ and so forth; it is a simple question of counting balls that give rise to each event. If you want to make that choice of each bowl equally likely, then you can achieve this, but only by modifying the probability measure, or even the probability space. For instance one could envision three independent uniform drawings: one of a bowl, one of a ball from the first bowl, and one of a ball from the second bowl; together this gives $$2\times7\times9=126$$ points for $$\Omega$$, for which one could decree a uniform measure. For a point in this new $$\Omega$$ one could associate the outcome of selecting the chosen ball from the chosen bowl (this set $$\Omega$$ is larger than needed, but that is not a problem as long as every element of $$\Omega$$ determines a unique outcome). With this setup each event of choosing one particular ball from the first bowl has probability $$\frac9{126}=\frac1{14}$$, and for each ball in the second bowl it is $$\frac7{126}=\frac1{14}$$; the probability of the even of choosing some white ball is $$3\times\frac9{126}+4\times\frac7{126}=\frac{55}{126}$$. One could modify the second scenario by regrouping together in $$\Omega$$ all points that lead to the choice of the same ball, giving rise to the original set $$\Omega$$ of $$16$$ points but with in this case a non-uniform probability distribution. In any case a well posed probability problem should be clear about the probability distribution used.