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Probability Formula Question
#1
I know and understand that:

P(B|A) = P(AandB)/P(A)

but does...

P(A|B) always = (PAandB)/P(B) as well?

Thanks!
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#2
AND...is this ALWAYS true regardless of whether the events are dependent or independent?

P(AandB) = P(A|B) * P(B) = P(B|A) * P(A)
We have a great group of "academic counselors" on this board. You will never find a greater group of supporters willing to offer you the benefit of their experience, helpful advice, and constructive criticism.

Don't miss out on something great just because it might also be difficult.

Road traveled: AA > BS > MSPM (2016)

If God hadn't been there for me, I never would have made it. Psalm 94:16-19
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#3
Wait, I think I get it.

If the events are independent, then P(AandB) = P(A) * P(B)

If the events are dependent, then P(AandB) = P(A/B) * P(B)

Is that right? I sure hope so. Sad
We have a great group of "academic counselors" on this board. You will never find a greater group of supporters willing to offer you the benefit of their experience, helpful advice, and constructive criticism.

Don't miss out on something great just because it might also be difficult.

Road traveled: AA > BS > MSPM (2016)

If God hadn't been there for me, I never would have made it. Psalm 94:16-19
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#4
Quote:P(B|A) = P(AandB)/P(A)

P(A|B) always = P(AandB)/P(B)

This is correct. The numerator is always P(AandB) for Bayes Theorem. The denominator is the probability which the outcome depends on (tip: it's always the alphabet on the right hand).

P(A|B) can be defined as the probability of A happening on the condition that B happens.
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#5
In regards to your first post, keep in mind that A and B are just random variables. They could represent any number and thus are interchangable
Edit: I just realized that this post was from over a month ago. Whoops! thought it was only a few days old
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