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I've got to show that:

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

I'm not sure how to get to this.

Surely the probability of A occurring given A and B occurs is 1?

Or, by the equation...

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

Thanks for your help

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3 Answers 3

Unless I'm missing something, you're completely correct, and the question is wrong.

$$\mathbb P(A|A\cap B) \equiv \frac{\mathbb P(A\cap (A\cap B))}{\mathbb P(A\cap B)} = \frac{\mathbb P(A\cap B)}{\mathbb P(A\cap B)} = 1$$

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Yet another explanation is that the typo(s) in question are that $\cup$ was replaced by $\cap$ (or an u in \cup got changed to an a to give \cap for it is certainly true that $$P(A \mid A\cup B) = \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A \cup B)}.$$

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I think, it must be a typo. It is also well possible that $\Bbb P(A)>\Bbb P(A\cap B)$, so the right hand side is $>1$.

The similar equation what we have is $$\Bbb P(A\cap B\mid A)=\frac{\Bbb P(A\cap B)}{\Bbb P(A)}\,.$$

Probably the author made a correction note to this exercise to 'exchange the terms', but later when he executed these notes, exchanged the terms in the fractions..

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