Imagine two trials that are not necessarily independent. The first succeeds with probability p1, the second succeeds with probability p2, and both trials succeed with probability p3. If the results are independent, p3 = p1×p2, but this need not be the case.
Bayes (biography) demonstrated the following connection between the two events. If we know the first outcome is true, the second is also true with probability p3/p1. This is clear when the events are independent, and p3 = p1p2.
Again, return to the model of rational numbers, and generalize to the reals by continuity. There are t total outcomes for both experiments. Let s1 be the number of possibilities that cause the first test to succeed, while the second succeeds in s2 cases. Finally s3 possibilities produce two true results. As sets, s3 = s1∩s2. As numbers, s3 ≤ s1 and s2. Restrict attention to s1, since we know the first test was a success. The second outcome is true with probability s3/s1. This is the same as s3/t over s1/t, or p3/p1.
Let's consider an example. Let one box contain two black marbles while another box has a black marble and a white marble. Select a box at random and withdraw a black marble. What are the odds that you chose box number one?
Here, p1 (black marble) is 3/4, p2 (first box) is 1/2, and p3 (both) is 1/2. These are not independent events, since 3/4*1/2 = 3/8, which is not equal to 1/2. We know the marble is black, so divide p3 by p1 and get 2/3. We have the first box two out of three times. This makes sense, since two of the three black marbles live in the first box.
Note that the conditional probability of the second event, based on the knowledge that the first result is true, can be computed without knowing the unconditional probability p2. That's good, since we often don't know p2.