My Favorite One-Liners: Part 26

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

Here’s a problem that could appear early in a probability class:

Let P(A) = 0.2, P(B) = 0.4, and P(A \cup B) = 0.5. Find P(A \mid B).

The standard technique for solving this problem involves first finding P(A \cap B) using the Addition Rule:

P(A \cup B) = P(A) + P(B) - P(A \cap B)

0.5 = 0.2 + 0.4 - P(A \cap B)

P(A \cap B) = 0.1

From here, the Multiplication Rule can be used (or, equivalently, the definition of a conditional probability):

P(B \cap A) = P(B) \cdot P(A \mid B)

0.1 = 0.4 P(A \mid B)

0.25 = P(A \mid B)

So far, so good.

Now let me add a small twist to the original problem that creates a small difficulty when solving:

Let P(A) = 0.2, P(B) = 0.4, and P(A \cup B) = 0.5. Find P(A \cap B \mid A \cup B).

Proceeding as before, we obtain

P( [A \cap B] \cup [A \cup B] ) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)

The value of $P(A \cup B)$ is obvious. But how do we evaluate the left side?

If I’m teaching an advanced probability class, I might expect them to use DeMorgan’s Laws. However, it’s a whole lot easier to reason out the left hand side: I’m looking for the probability that both A and B happen or else at least one of A and B happen. Well, that’s clearly redundant: if both A and B happen, then certainly at least one of A and B happen.

Here’s my one-liner, which I say, if possible, using only one breath of air:

Clearly, this is redundant. It’s like saying Dr. Q is my professor and he’s a total stud. It’s redundant. It’s obvious. There’s no need to actually say it.

After the laughter settles from this bit of braggadocio, the A \cup B can be safely dropped from the left side:

P( A \cap B) = P(A \cup B) \cdot P(A \cap B \mid A \cup B)

0.1 = 0.5 \cdot P(A \cap B \mid A \cup B)

0.2 = P(A \cap B \mid A \cup B)

However, I need to emphasize that dropping the term on the left side is a special feature of this particular problem since one set was a subset of the other, and that students shouldn’t expect to always be able to do this when computing conditional probabilities.

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