Question

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)?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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.

Answer 5

\[ P(D|E) = {P(D\cap E) \over P(E)} \implies P(D\cap E) = P(D|E) \times P(E) \]

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.

Answer 6

It's clear? or you need more help?

Answer 7

2 seconds, doing it on paper

Answer 8

Sure :)

Answer 9

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

Answer 10

http://en.wikipedia.org/wiki/Conditional_probability#Definition

Answer 11

Ok, I see it now, thanks!

Answer 12

Glad to help :)