## anonymous 4 years ago Probablity question: $P(D|E) = P(DE) div P(E)$ $P(E|D)P(D) \div P(E|D)P(D) + P(E|D^{c})P(D^{c})$ On the second line, I'm not sure 100% how we arrive at P(DE) = P(E|D)P(D). Can someone confirm it's because it's the contrapositive of P(D|E)P(E)?

1. anonymous

$P(D|E) = {P(D\cap E) \over P(E)}$

2. anonymous

$P(D|E) = P(DE) \div P(E) = P(E|D)P(D) \div P(E|D)P(D) + P(E|D^{c})P(D^{c})$

3. anonymous

This seems latex intensive, I am scared to answer it without edit feature, sorry.

4. anonymous

Then, what's the name of the rule/law that states that P(DE) = P(E|D)P(D)? I understand it equals P(D|E)P(E) but not he latter.

5. anonymous

$P(D|E) = {P(D\cap E) \over P(E)} \implies P(D\cap E) = P(D|E) \times P(E)$ again, $P(E|D) = {P(D\cap E) \over P(D)} \implies P(D\cap E) = P(E\cap D) = P(E|D) \times P(D)$ Hopefully, it's clear now.

6. anonymous

It's clear? or you need more help?

7. anonymous

2 seconds, doing it on paper

8. anonymous

Sure :)

9. anonymous

To be somewhat pedantic, $P(A|B) \triangleq \frac{P(A \cap B)}{P(B)}$

10. anonymous
11. anonymous

Ok, I see it now, thanks!

12. anonymous