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

Road traveled: AA (2013) > BS (2014) > MS (2016) > Doctorate (2024)

If God hadn't been there for me, I never would have made it. Psalm 94:16-19
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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)
Don't miss out on something great just because it might also be difficult.

Road traveled: AA (2013) > BS (2014) > MS (2016) > Doctorate (2024)

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

Road traveled: AA (2013) > BS (2014) > MS (2016) > Doctorate (2024)

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