A box contains 6 red balls, 4 blue balls and 2 green

Solution

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