If a pair of fair dice is cast, the probability that the sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is given by:
[tex] \frac{P(sum\, of\, 9\, \cup\,3)}{P(3)} [/tex]
The number of outcomes of sum of 9 where at last one is 3 is (3, 6) and (6, 3) = 2.
The number of outcomes of last one of the numbers falling uppermost is a 3 is (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3) and (6, 3) = 11.
Therefore, the
probabiliy that the sum of the numbers falling uppermost is 9, given
that at least one of the numbers falling uppermost is a 3 is 2 / 11.
