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 , , and . Find .

The standard technique for solving this problem involves first finding using the Addition Rule:

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

So far, so good.

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

Let , , and . Find .

Proceeding as before, we obtain

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 and happen or else at least one of and happen. Well, that’s clearly redundant: if both and happen, then certainly at least one of and 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 can be safely dropped from the left side:

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.