I know that p(a,b) = p(a|b)p(b) = p(b|a)p(a). But what is

pachaquis3s

pachaquis3s

Answered question

2022-06-20

I know that p(a,b) = p(a|b)p(b) = p(b|a)p(a).
But what is the reason for this?
Why don't p(a,b) = p(a|b)p(b|a) since they all involve with each other.

Answer & Explanation

Bruno Hughes

Bruno Hughes

Beginner2022-06-21Added 24 answers

It is easy to show counter-examples that Pr ( A , B ) need not be Pr ( A | B ) Pr ( B | A ): for example, if A and B always happen together, but do not always happen, then Pr ( A | B ) = Pr ( B | A ) = 1 but Pr ( A , B ) < 1.

Now for the conditional probability statement, let's write A c to be the complement of A etc. There are four mutually exclusive possible outcomes: A B; A c B; A B c ; and A c B c . You have
Pr ( B ) = Pr ( A B ) + Pr ( A c B ) Pr ( A B ) + Pr ( A c B c ) + Pr ( A B ) + Pr ( A c B c )
where the denominator sums to 1, and also have
Pr ( A | B ) = Pr ( A B ) Pr ( A B ) + Pr ( A c B )
so long as the denominator is not 0, so multiplying them together gives
Pr ( A | B ) Pr ( B ) = Pr ( A B ) .
A similar argument gives Pr ( B | A ) Pr ( A ) = Pr ( A B ) .

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school probability

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?