Hope $A \cap B$ occurs

Question

I am studying for some interviews and came across this problem from Blitzstein's book.

Let A and B be events with $0 < P(A \cap B) < P(A) < P(B) < P(A \cup B) < 1$. You are hoping that both $A$ and $B$ occurred. Which of the following pieces of information would you be happiest to observe: that $A$ occurred, that $B$ occurred, or that $A \cup B$ occurred?

The problem seems pretty trivial once you do the math:

$P(A \cap B | A \cup B)$

= $\frac{P(A \cup B | A \cap B)P(A \cap B)}{P(A \cup B)}$

= $\frac{P(A \cap B)}{P(A \cup B)}$

Similarly for $P(A \cap B | A)$ and $P(A \cap B | B)$, we arrive at $\frac{P(A \cap B)}{P(A)}$ and $\frac{P(A \cap B)}{P(B)}$, respectively.

Because $P(A)$ is the smallest, $P(A \cap B | A)$ is the largest and our answer is $A$.

Generally, this makes sense to me as if we want two events to occur and the less likely one will happen, it is ideal. Now what doesn't make sense to me is why $A \cup B$ makes $A \cap B$ less likely than just $B$.

Intuitively, it feels like if we have two events, it is less informative that the less likely one occurred than one of the two arbitrarily occurred. The idea that $A \cup B$ has the potential for $A$ to occur makes it feel like it should make the $P(A \cap B)$ more likely.

Could someone either point out a flaw in my math which would make my intuition correct, or critique my intuition? Thanks.

Answer 1

We are given that $P(A)\lt P(B)\lt P(A\cup B)$.

This means that there are fewer ways for $A$ to occur than $B$ and fewer ways for $B$ to occur than $A\cup B$.

When we condition on an event, we reduce the probability space. For example, the probability of getting a two when we roll a six sided die is $\frac{1}{6}$ because there is one favorable outcome in a probability space of six outcomes. If, instead, we ask for the probability to get a two given that the outcome was even, our probability is greater because, although we still have only one way to get a two, there are now only three outcomes in our probability space, viz $2,4,6$, so our probability increases from $\frac{1}{6}$ to $\frac{1}{3}$.

If we say $A$ occurred, we only consider those outcomes in which $A$ occurred and we can eliminate any outcomes where $A$ didn't occur. Likewise, if $B$ occurred, we can eliminate any outcomes where $B$ didn't occur. But if $A\cup B$ occurred we cannot eliminate outcomes where $A$ didn't occur or $B$ didn't occur because we can't be sure which one actually occurred.

Therefore, the probability space when $A\cup B$ occurs is greater than when we know which of $A$ and $B$ occurred so the probability that $A\cap B$ occurred is less.

Answer 2

First of all, wonderful question! Intuition should be critiqued in order to be trusted! Now, onto the question.

Why $A∪B$ makes $A∩B$ less likely than just $B$ ?

If you observe only $B$, there are two outcomes - $B\setminus A$ and $A \cap B$. So roughly speaking in only $1$ of the $2$ scenarious you get the desired outcome.

Now, suppose you observe $A \cup B$. Three outcomes are possible - $B\setminus A$, $A\setminus B$ and $A \cap B$. Out of these, the desired outcome is the last one so only $1$ out of the $3$ will suit you.

We're talking in terms of intuition, so since $1/2>1/3$, you should be happier to observe $B$, than $A \cup B$.