A box contains 6 red balls, 4 blue balls and 2 green balls. Two balls are drawn at random from the box without replacement. If it is known that at least one of the drawn balls is red, what is the probability that both the drawn balls are red?

ATQ, Total balls = 6 + 4 + 2 = 12 We draw 2 balls without replacement. Let: A = event “both balls are red” B = event “at least one ball is red” We need P(A | B) = P(A ∩ B) / P(B). But if both balls are red, then certainly at least one is red ⇒ A ⊂ B. So P(A ∩ B) = P(A). First compute: Total number of ways to choose 2 balls from 12 = C(12, 2) = (12×11)/2 = 66 P(A): both balls red Number of ways to choose 2 red balls from 6 = C(6, 2) = (6×5)/2 = 15 So P(A) = 15 / 66 P(B): at least one ball red Easier to find complement: “no red ball” (both balls are non-red). Non-red balls = blue + green = 4 + 2 = 6 Ways to choose 2 balls from these 6 = C(6, 2) = 15 So P(no red) = 15 / 66 Therefore P(at least one red) = 1 − 15/66 = (66 − 15)/66 = 51/66 Now: P(A | B) = P(A) / P(B) = (15/66) ÷ (51/66) = 15/51 = 5/17