What is the intersection of an event with itself?

Question

The probability of the intersection of two event is:

$P(A \cap B) = P(A)P(B)$

If the two events are the same, i.e

$P(A \cap A) = P(A)P(A) = P(A)^2$

However, the logic tells us that the probability of an event intersecting itself is 1, since it is contained within itself.

So the probability must be $1$ or $P(A)^2$

Answer 1

You only have that:

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

if $A$ and $B$ are independent events ... but $A$ and $A$ are clearly not.

However, what is always true is that

$P(A \cap B) = P(A) \cdot P(B|A)$

Thus, you have that:

$P(A\cap A)=P(A) \cdot P(A|A)$

But, obviously we have that $P(A|A)=1$: the probability that $A$ happens given that $A$ happens is $1$

Hence:

$P(A \cap A) = P(A) \cdot P(A|A)= P(A) \cdot 1 = P(A)$

Of course, this make total sense, since set-theoretically, we immediately have that $A \cap A=A$, and thus we could have immediately said that $P(A \cap A) =P(A)$. But it is nice to confirm that it works out using the general formula as well. Sanity check! :)

Answer 2

$P(A \cap B) = P(A)P(B)$ only if $A$ and $B$ are independent. $A$ and $A$ can hardly be so.

$P(A \cap B) = P(A)P(B|A)$ and for $B=A$, $P(A \cap A) = P(A)P(A|A)=P(A)$, which is not much of an info.

Answer 3

The stated "rule" only holds when the events are independent. $A$ is not independent from itself.

Answer 4

$P(A \cap B) = P(A)P(B)$ is not always true, only if $A$ and $B$ are independent.

Clearly an event is not (usually) independent of itself.

Also, it is not true that "the probability of an event intersecting itself is $1$", $P(A\cap A)=1$. What you presumably mean is $P(A\cap A)=P(A)$.