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)?

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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)?

Mathematics
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\[ P(D|E) = {P(D\cap E) \over P(E)} \]
\[P(D|E) = P(DE) \div P(E) = P(E|D)P(D) \div P(E|D)P(D) + P(E|D^{c})P(D^{c})\]
This seems latex intensive, I am scared to answer it without edit feature, sorry.

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Other answers:

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.
\[ P(D|E) = {P(D\cap E) \over P(E)} \implies P(D\cap E) = P(D|E) \times P(E) \] 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.
It's clear? or you need more help?
2 seconds, doing it on paper
Sure :)
To be somewhat pedantic, \[ P(A|B) \triangleq \frac{P(A \cap B)}{P(B)} \]
http://en.wikipedia.org/wiki/Conditional_probability#Definition
Ok, I see it now, thanks!
Glad to help :)

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