# What is the intersection of an event with itself?

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 $$P(A)^2$$

• A intersect A is A. It is just P(A) – randomgirl Apr 18 at 1:02

## 4 Answers

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! :)

• Thanks for your answer – Mattiu Apr 18 at 1:19
• @Mattiu You're welcome! Remember that more general formula, because that is how you derive Bayes' Law as well! – Bram28 Apr 18 at 1:21
• Good answer. ${}{}$ – Randall Apr 18 at 1:22

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

• Thanks for $P(A \cap A) = P(A)P(A|A)=P(A)$ – Mattiu Apr 18 at 1:13

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

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