Example

Suppose that somebody secretly rolls two fair six-sided dice, and we wish to compute the probability that the face-up value of the first one is 2, given the information that their sum is no greater than 5.

  • Let D1 be the value rolled on die 1.
  • Let D2 be the value rolled on die 2.


Probability that D1 = 2

Table 1 shows the sample space of 36 combinations of rolled values of the two dice, each of which occurs with probability 1/36, with the numbers displayed in the red and dark gray cells being D1 + D2.

D1 = 2 in exactly 6 of the 36 outcomes; thus P(D1 = 2) = 6⁄36 = 1⁄6:

Table 1

+ D2
1 2 3 4 5 6
D1 1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

Probability that D1 + D2 ≤ 5

Table 2 shows that D1 + D2 ≤ 5 for exactly 10 of the 36 outcomes, thus P(D1 + D2 ≤ 5) = 10⁄36:

Table 2

+ D2
1 2 3 4 5 6
D1 1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12


Probability that D1 = 2 given that D1 + D2 ≤ 5

Table 3 shows that for 3 of these 10 outcomes, D1 = 2.

Thus, the conditional probability P(D1 = 2 | D1+D2 ≤ 5) = 3⁄10 = 0.3:


Table 3

+ D2
1 2 3 4 5 6
D1 1 2 3 4 5 6 7
2 3 4 5 6 7 8
3 4 5 6 7 8 9
4 5 6 7 8 9 10
5 6 7 8 9 10 11
6 7 8 9 10 11 12

Here, in the earlier notation for the definition of conditional probability, the conditioning event B is that D1 + D2 ≤ 5, and the event A is D1 = 2. We have P(A\mid B)={\tfrac {P(A\cap B)}{P(B)}}={\tfrac {3/36}{10/36}}={\tfrac {3}{10}}, as seen in the table.