Pr (A \ B) = The probability of A, given that B has occurred:
Pr (B \ A) * Pr (A)
Pr (B \ A) * Pr (A) + Pr (B \ not A) * Pr (not A)
*
To illustrate the idea, this example is taken from 'IntroSTAT' by Les Underhill and Dave Bradfield:
You feel ill at night and stumble into the bathroom, grab one of three bottles in the dark and take a pill. An hour later you feel really ghastly, and you remember that one of the bottles contains poison and the other two aspirin.
Your handy medical text says that 80% of people who take the poison will show the same symptoms as you are showing, and that 5% of people taking aspirin will have them.
Let B be the event 'having the symptoms'
A be the event 'taking the poison'
Then 'not A' is the event 'taking aspirin'
What is the probability that you took the poison given that you have got the symptoms? i.e. what is Pr (A \ B)
From the information supplied by the handy medical text
Pr (B \ A) = 0.8 and Pr (B \ not A) = 0.05
From our groping round in the dark, we conclude that
Pr (A) = 1/3 and Pr(not A) = 2/3
Thus Pr (A \ B)
= {0.8 * 1/3} divided by {0.8 * 1/3 + 0.05 * 2/3} = 0.89
We recommend that you call the doctor!
