**Question **: A pair of unbiased dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is

(A) 2/5

(B) 3/5

(C) 4/5

(D) none of these

**Answer **: option (A)

**Solution **:

Let A denote the event that the sum of 5 occurs, B the event that a sum of 7 occurs and C the event that either a sum of 5 nor a sum of 7 occurs. We have,

P(A) = 4/36 = 1/9, P(B) = 6/36 = 1/6, P(C) = 26/36 = 13/18,

Thus the probability that A occurs before B is,

=P[A or (C n A) P(C n C n A) or ….]

=P(A) + P(C n A) + P(C n C n A) +…

=P(A) + P(C) P(A) + P\left( C\right) ^{2}P\left( A\right) +….

=\dfrac{1}{9}+\left( \dfrac{13}{18}\right) \times \left( \dfrac{1}{9}\right) +\left( \dfrac{13}{18}\right) ^{2}\times \left( \dfrac{1}{9}\right) +….

=\dfrac{\dfrac{1}{9}}{1-\dfrac{13}{18}}

=2/5