0
$\begingroup$

I am stuck on a vital step in this process, and that is about the conditional probability aspect of it. Once I get that step, the integration and using the gamma function should go fine.

So my question is the following: Will P[x > 4 | x > 2] = P[x > 4 and x > 2]/P[x>2] = P[x>4] / P[x>2] OR will it be P[2< x<4]/P[x>2]? The first result gives me a really small probabiliy of 0.0557 and the latter one gives me 0.94, which are two entirely different answers. Thanks in advance.

Gamma function

$\endgroup$
0
$\begingroup$

By definition of conditional probability, $P(A|B) = P(A \ \text{and}\ B)/P(B)$.

$\endgroup$
  • $\begingroup$ Right. So will it be that x has to be great than 4 because that is the only intersection between those two? $\endgroup$ – Sander Heieren Oct 16 '18 at 3:17
  • $\begingroup$ "$x > 4$ and $x > 2$" is equivalent to $x > 4$. $\endgroup$ – Robert Israel Oct 16 '18 at 3:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.