Bayes Formula derivation inconsistency with conditional probability formula

[andresap.me]

Related to the derivation of Bayes theorem in Week 1 - Bayes Theorem Mathematical Formula.

I’m having trouble getting past the derivation of Bayes theorem.

The problem: the formula for conditional probability is not used consistently within the video, and in relation to the lesson on conditional probabilities where it is initially introduced.

Here’s the context: With events A: Being sick and B: diagnosed sick, we want to get P(A|B), namely the probability of being sick given that we were diagnosed sick.

The first step is to use the conditional probability formula to find an expression for P(A|B), which is what I’m struggling with.

In the lecture, the lecturer gets P(A|B)=P(A∩B)/P(B) from

the conditional probability formula P(A | B) P(B) = P(A∩B). So if we divide by P(B) we get that P(A|B)=P(A ∩ B)/ P(B).
(copied from the transcript)

But this is, to my understanding, not consistent with the lesson on conditional probabilities, since the formula would be

P(A∩B) = P(A) P(B|A),
or P(B∩A)=P(B) P(A|B).

It makes sense to me that the probability of being sick given that I’m diagnosed sick is the population of those who are sick AND have been diagnosed sick out of the total population who have been diagnosed sick. This is what the video on the Bayes Theorem implies, but it’s not what the formula for the conditional probability says (P(A∩B)=P(A) P(B|A)).

A little later in the video, the formula for conditional probability is used again, this time consistently as P(A∩B)=P(A) P(B|A).

I’d find some clarification on this very useful!

Thank you!

[pastorsoto]

Hi @andresap.me great question, let’s try to go step by step

1. Start with the basics of conditional probability:

   • The probability of A happening given B is:
     P(A | B) = P(A ∩ B) / P(B)
     (This means: “Out of all the times B happens, how often does A also happen?”)

2. Rewriting the joint probability (P(A ∩ B)):

   • You can also write P(A ∩ B) as:
     P(A ∩ B) = P(A) × P(B | A)
     (This means: “The chance of A happening, multiplied by the chance of B happening if A is true.”)

3. Combine these ideas to get Bayes’ Theorem:

   • Rearrange the formulas to find P(A | B):
     P(A | B) = [P(A) × P(B | A)] / P(B)
     (This tells you: “The chance of A given B depends on how likely A is overall, how likely B is if A happens, and how likely B is overall.”)

If you’re trying to figure out if someone is sick (A) based on a positive test result (B). Bayes’ Theorem helps you combine:

• How common the sickness is (P(A)),
• How accurate the test is (P(B | A)),
• And how often the test gives positive results overall (P(B)).

So, there are just different ways of putting the formula but it means the same

I hope this helps!

[andresap.me]

Okay thank you for your reply! I think it does help.

To me it makes sense from conditional probability P(A | B) = P(A ∩ B) / P(B).

I think I got confused because I thought we were trying to obtain an expression for P(A|B) using the formula for P(A∩B), which in this case didn’t quite add up. By looking at it as two separate expressions, or ideas like you explained it, I can move on in peace now.

Thank you!