# 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.