The probability of 7 when rolling two die is 1/6 (= 6/36) because the sample space consists of 36 equiprobable elementary outcomes of which 6 are favorable to the event of getting 7 as the sum of two die. Denote this event A: P(A) = 1/6.

Consider another event B which is having at least one 2. There are still 36 elementary outcomes of which 11 are favorable to B; therefore, P(B) = 11/36. We do not know whether B happens or not, but this is a legitimate question to inquire as to what happens if it does. More specifically, what happens to the probability of A under the assumption that B took place?

The assumption that B took place reduces the set of possible outcomes to 11. Of these, only two - 25 and 52 - are favorable to A. Since this is reasonable to assume that the 11 elementary outcomes are equiprobable, the probability of A under the assumption that B took place equals 2/11. This probability is denoted P(A|B) - the probability of A assuming B: P(A|B) = 2/11. More explicitly P(A|B) is called the conditional probability of A assuming B. Of course, for any event A, P(A) = P(A|Ω), where, by convention, Ω is the universal event - the whole of the sample space - for which all available elementary outcomes are favorable.

We see that, in our example, P(A|B) ≠ P(A). In general, this may or may not be so.

Retracing the steps in the example,
P(A|B) = P(A∩B) / P(B)

Since there are 36 possible outcomes when we roll two die, we have that P(A and B) = 2/36

P(B) = 11/36

P(A|B) = (2/36)/(11/36)
P(A|B) = 2/11

In general, if A and B are independent events, then we have
P(A|B) = P(A)