A bag contains 3 red, 4 blue, and 5 green balls. Three balls are drawn from the bag one by one without replacement. If it is given that the third ball drawn is red, the probability that the first two balls drawn are of different colours is p q , where p and q are coprime positive integers. The value of p + q is :

Step-by-step solution

Let R_i, B_i, G_i denote the events that the i^ th ball drawn is red, blue, or green respectively. The total number of balls is 12 . By symmetry, the probability that the third ball is red is P(R₃) = 3 12 = 1 4 . We need to find P( Different _ 1,2 | R₃) = 1 - P( Same _ 1,2 | R₃) . The event that the first two balls are of the same colour and the third is red is given by (R₁ R₂ R₃) (B₁ B₂ R₃) (G₁ G₂ R₃) . P(R₁ R₂ R₃) = 3 12 2 11 1 10 = 6 1320 P(B₁ B₂ R₃) = 4 12 3 11 3 10 = 36 1320 P(G₁ G₂ R₃) = 5 12 4 11 3 10 = 60 1