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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ P(DE) = {P(D\cap E) \over P(E)} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[P(DE) = P(DE) \div P(E) = P(ED)P(D) \div P(ED)P(D) + P(ED^{c})P(D^{c})\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This seems latex intensive, I am scared to answer it without edit feature, sorry.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then, what's the name of the rule/law that states that P(DE) = P(ED)P(D)? I understand it equals P(DE)P(E) but not he latter.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ P(DE) = {P(D\cap E) \over P(E)} \implies P(D\cap E) = P(DE) \times P(E) \] again, \[ P(ED) = {P(D\cap E) \over P(D)} \implies P(D\cap E) = P(E\cap D) = P(ED) \times P(D) \] Hopefully, it's clear now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's clear? or you need more help?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.02 seconds, doing it on paper

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0To be somewhat pedantic, \[ P(AB) \triangleq \frac{P(A \cap B)}{P(B)} \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://en.wikipedia.org/wiki/Conditional_probability#Definition

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ok, I see it now, thanks!
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