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The two events E and F have probabilities of $P(E) = P(F) = 0.5$ and they are dependent since $P(E| \lnot F) = 0.6$. What's $P(E|F)$?

I know Bayes theorem, it's just that I don't exactly know what $P(E \cap F)$ is.

And with relation to that, I have another question:

Considering we know $P(E|\lnot F)$, what other $P$'s can we extract from that?

I'm sorry this is an easy question, but it's mainly just to get me started.

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You know that $P(E\mid\lnot F)=0.6$. You also know that $P(E\mid\lnot F)=\frac{P(E\;\&\;\lnot F)}{P(\lnot F)}\;,$ and you know that $P(F)=0.5$. Can you put the pieces together to find $P(E\;\&\;\lnot F)$ and then use your knowledge of $P(E)$ to get $P(E\;\&\,F)$?

If you work your way through this, you’ll get a good idea of the answer to your second question as well.

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    @Sorin: Yes, it would: the occurrence of $F$ would not have affected the likelihood of $E$.2012-05-13
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Proof of the fact you're interested in:

\begin{align*} \Pr(E) &= \Pr(E\cap(F\cup\lnot F)) \\ &= \Pr((E\cap F)\cup(E\cap\lnot F)) \\ &= \Pr(E\cap F) + \Pr(E\cap\lnot F) \end{align*}

The last equality is justified because:

$ (E\cap F)\cap(E\cap\lnot F) = E\cap(F\cap\lnot F) = E\cap\emptyset = \emptyset $