The probability of an event occurring given that another event has already occurred is called a conditional probability. We will explain this using examples.

There are four balls in a box, marked with numbers: two black balls with numbers 1, 2 and two white balls with numbers 3 and 4. We draw two balls without replacement (i.e. the ball drawn first is put aside and is not used in the second sampling). The sample space is {(x,y): x ≠ y and x,y = 1, 2, 3, 4}. Hence | Ω| = 4*3 = 12. We assume that the elementary events are equally probable, so the probability of drawing a pair (x,y) is equal for all x and y, P(x,y) = 1/12.

Let's consider the events: A = "white ball in the second draw", B = "black ball in the first draw". There are six events favorable to the event A: (1,3), (1,4), (2,3), (2,4), (3,4), (4,3), so P(A) = 6/12. There are also six events favorable to the event B: (1,2), (1,3), (1,4), (2,1), (2,3), (2,4).

If we know that the event B has happened after the first draw, what is the probability of the event A? It's obvious that knowing that the first ball is black, we know that in the second draw we choose from two white balls and a black one, i.e. balls 1, 3, 4 or 2, 3, 4. In this new situation, the set of possible events in the second draw is a black ball or the white ball number 3 or the white ball number 4. There are only three possibilities. As two of them are favorable to the event A, the probability of the event A, assuming that the event B has happened, is 2/3.

Definition 13.6.1

The probability of the event A on condition that the event B has happened, designated by P(A|B), is expressed by the formula:

P(A|B) = P(A ∩ B)/ P(B) if P(B) >0.

Let's consider once again the situation described in the example 13.6.1. The probability of the event A ∩ B is 4/12, and the probability of the event B is 6/12. Hence, according to the definition 13.6.1, P(A|B) = 4/6 = 2/3.

We roll 3 different dice. Let A = "1 received at least on one die" and B = "a different number received on each die". What is the probability of the event A assuming that the event B has happened?
Solution. The elementary events in this trial are triples (x,y,z), where x, y, z = 1, 2, 3, 4, 5, 6. There are 6*6*6 of them. There are as many events favorable to the event B as there are one-to-one functions defined in the three-element set and with the values in the set of 6 elements. There are 6*5*4 of them, see section 11.2. So P(B) = 6*5*4/(6*6*6). The events favorable to the event A ∩ B are triples (1,x,y) ,(x,1,y), (x,y,1), where x is one of 5 values (as it can't be 1), and y is one of 4 values (it has to be different from x and from 1). P(A ∩ B) = 5*4*3/63. Finally

Example 13.6.3
We roll a die twice. What is the probability of receiving three pips, if you know that 1 has appeared in the first roll.

Solution. The sample space when rolling a die twice consists of 36 elements. The probability of receiving one pip in the first roll is 1/6, and the probability of receiving one pip in the first roll and two pips in the second roll is 1/36. Hence, the probability in question is (1/36)/(1/6), i.e. 1/6.

Question 13.6.1: In the group of 100 people there are 19 male and 12 female that are smokers. What is the probability that a randomly selected smoker is male?