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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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2 Answers 2

up vote 1 down vote accepted

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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Well, 0.6 = P(E & not F) / 0.5. Thus, P(E & not F) is 0.3. Is P(E&F) + P(E&notF) meant to be 0.5? In that case, P(E&F) = 0.2, so P(E|F) = 0.4? –  Sorin Cioban May 13 '12 at 20:34
    
@Sorin: Yes, $P(E)=P(E\;\&\,F)+P(E\;\&\;\lnot F)$, and the rest looks good. And as you can see, you have access to every probability involving some combination of $E$ and $F$. –  Brian M. Scott May 13 '12 at 20:37
    
I see. Thanks, I appreciate your help :) By the way, if they were independent, it would've been plain and simply 0.5, correct? –  Sorin Cioban May 13 '12 at 20:40
    
@Sorin: Yes, it would: the occurrence of $F$ would not have affected the likelihood of $E$. –  Brian M. Scott May 13 '12 at 20:42

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:

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

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