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Let $A$ and $B$ be events, not necessarily independent.

The way I was taught was $\Pr(A \land B) = \Pr(A)\cdot\Pr(B|A)$. This is to say the probability of $A$ and then $B$ is equal to the probability of $A$ multiplied by the probability of $B$ given $A$ has already occured. This makes intuitive sense to me because it states that $A$ occurs before $B$ therefore we can use $A$ to determine the impact on $B$.

However, I see online that the equation can be switched to $\Pr(A \land B) = \Pr(B)\cdot\Pr(A|B)$, without switching to $\Pr(B \land A)$. Is it safe to say that $\Pr(A \land B) = \Pr(B \land A)$?

Lucas Resende
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Shawn
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1 Answers1

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Yes, because as sets $A \cap B = B \cap A$. (And really, that's what probability functions measure: $P(A)$ is essentially "the proportion of our sample space which $A$ occupies.")

Note that $A \cap B$ is the set of all elements which are in both $A$ and $B$. Semantically, there's no distinction if we were to switch $B$ and $A$ around after all. (You can formally prove this if you desire, see below.) So it makes even more sense that $P(A \cap B) = P(B \cap A)$: they consist of the same elements, so the probabilities should be the same.

Proof $A \cap B = B \cap A$:

In set theory we say two sets $S,T$ are equal if we have $S \subseteq T$ and $T \subseteq S$. That is, whenever $x \in S$, we show $x \in T$ (and vice versa). We also recall the definition of intersection:

$$S \cap T = \{ x \mid x \in S \text{ and } x \in T \}$$

So, then:

$$\begin{align} x \in A \cap B &\implies x \in A \text{ and } x \in B \\ &\implies x \in B \text{ and } x \in A \\ &\implies x \in B \cap A \\ &\implies A \cap B \subseteq B \cap A \\ x \in B \cap A &\implies x \in B \text{ and } x \in A \\ &\implies x \in A \text{ and } x \in B \\ &\implies x \in A \cap B \\ &\implies B \cap A \subseteq A \cap B \end{align}$$

Since we have $A \cap B \subseteq B \cap A$ and $B \cap A \subseteq A \cap B$, then, $A \cap B = B \cap A$.

PrincessEev
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