Let E be an arbitrary event in a sample space S with P(E)> 0. The probability that an event A occurs once E has occurred or, in other words, the conditional probability of A given E, written P(A| E), is defined as follows:

P(A| E) = P(AnE)/P(E)

As seen in the adjoining Venn diagram, P(A| E) in a certain sense measures the relative probability of A with respect to the reduced space E.

In particular, if S is a finite equiprobable space and IAJdenotes the number of elements in an event A, then

P(AnE) = |A n E|/|S|, P(E)= |E|/|S| and so P(A|E) = P(AnE)/P(E) – |AnE|/|E|

That is,

Theorem: Let S be a finite equiprobable space with events A and E. Then P(A|E) =number of elements in A n E/number of elements in E or P(A|E) = number of ways A and E can occur/number of ways E can occur

### Multiplication Theorem for conditional probability

If we cross multiply the above equation defining conditional probability and use the fact that A n E = E n A, we obtain the following useful formula.

**Theorem:** P(En A) = P(E)P(A I E)

This theorem can be extended by induction as follows:

For any events A_{1},A_{2}, . . .,A_{n},

P(A_{1}n A_{2}n… nA_{n}) = P(A_{1})P(A_{2}|A_{1})P(A3|A_{1}nA_{2})***P(An|A_{1}nA_{2}n…nA_{n-1})

We now apply the above theorem which is called, appropriately, the multiplication theorem.

**Example:** A lot contains 12 items of which 4 are defective. Three items are drawn at random from the lot one after the other. Find the probability p that all three are nondefective.

The probability that the first item is non-defective is 8/12 since 8 of 12 items are nondefective.

If the first item is non-defective, then the probability that the next item is non-defective is 7/11 since only 7 of the remaining 11 items are nondefective.

If the first two items are non-defective, then the probability that the last item is non-defective is 6/10 since only 6 of the remaining 10 items are now nondefective.

Thus by the multiplication theorem,

P = 8/12 * 7/11 * 6/10 = 14/55