# Probability question involving intersections

1. Feb 20, 2015

### TheKracken

1. The problem statement, all variables and given/known data
and is in reference to the intersection of a set.
I know the formula that P(A and B ) = P(B) * P(A|B)

Now we are given something of the nature of P(A and B and C) I am trying to figure out a formula for this. From my searches on google I only find people giving the formula and I really would like to figure it out for my self.

Here is my attempt even though it does not appear to align with the correct formula.

P(A and B and C) = P((A and B) and (C)
= P(K and C ) ; such that P(A and B) = P(K)
= P(C) * P(K|C) ( now I need to do ( B|A)
= P(C) *P(A and B|C)*P(B|A)

What exactly am I doing wrong? I must have made some sort of logical error.

P(A)*P(C|A and B) *P(B|A)

2. Feb 20, 2015

If A happens first then B and finally C,
P(C/K) = P(C and K)/P(K)
=>P(C and K) = P(K)*P(C/K)
= P(A and B)*P(C/A and B)

P(A and B) = P(A)P(B/A) because A happens first and then B takes place.
Formula is correct. In your formula,B happens first.

3. Feb 20, 2015

### Ray Vickson

Note: this last expression is equal to
$$P(A)\, P(C|A \cap B)\, P(B | A) = P(C | A \cap B)\,\underbrace{ P(B|A)\,P(A)}_{=P(A \cap B)} \\ \hspace{2cm}= P(C | A \cap B) \,P(A \cap B) = P (C \cap A \cap B)$$
You can go backwards and "undo" $P(A \cap B \cap C) = P(C \cap A \cap B)$ to get the final result you want.

Last edited: Feb 20, 2015
4. Feb 21, 2015

### haruspex

The order of events is immaterial.