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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 $P(A)^2$

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  • $\begingroup$ A intersect A is A. It is just P(A) $\endgroup$ – randomgirl Apr 18 at 1:02
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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! :)

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

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  • $\begingroup$ Thanks for $P(A \cap A) = P(A)P(A|A)=P(A)$ $\endgroup$ – Mattiu Apr 18 at 1:13
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The stated "rule" only holds when the events are independent. $A$ is not independent from itself.

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

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