Is Bayes Theorem Correct for Calculating Probability of a Bomb?

[Math Is Hard]

This is a slide from a lecture I had:
http://www.geocities.com/thesquarerootoftwo/bayes.JPG
What we're looking at is a test for the presence of a bomb, like something airport screeners might use.
When the probability of the bomb, P(B) is figured, it looks like the calculation is done by taking the total number of times the bomb was present and dividing it by the total number of times the bomb was absent.

Is this correct? I was thinking you would take the total number of times the bomb was present and divide by (the total number of times absent + the total number of times present).

I have the same question about P(T) which I think is the probability of a positive test.

But I don't know jack about Bayes Theorem, so thanks for any help!

 

[Cyrus]

I am not sure what you're asking. But it seems like you are trying to divide by the total number of outcomes (all types of outcomes, positive/negative and bomb/nobomb.

The notation, P(B|T), means the probability of getting B (bomb), given that T(test) is positive.

So this means the total outcome set would be all of those that tested positve. You are only considering the probability realtive to everything that tested positive.

 

[Math Is Hard]

Hi Cyrus, I'm asking if the way P(B) and P(T) are calculated in the slide is correct.

 

[Cyrus]

Edit:

I think P(B) should be 1/1,000,101

and P(T) should be 101/1,000,101

Because the difference between 1,000,101 and 1,000,100 is small the answer will be the same. But if this were small numbers, the difference would be big.

I see what you mean now. I think it is wrong.

 

[Math Is Hard]

Thank you, Cyrus!