A box contains 4 red balls, 5 blue balls and 6 green balls. Three balls are drawn at random from the box without replacement. What is the probability that  exactly one  of the drawn balls is red,  given that  at least one of the drawn balls is red?

ATQ, Total balls = 4 + 5 + 6 = 15 Three balls drawn without replacement. We need: P(exactly 1 red | at least 1 red) By conditional probability: P(A | B) = P(A ∩ B) / P(B) Here, A: event “exactly one red is drawn” B: event “at least one red is drawn” Note: If exactly one red is drawn, then clearly at least one red is drawn, so A ⊂ B. Thus, P(A ∩ B) = P(A). So, P(exactly 1 red | at least 1 red) = P(exactly 1 red) / P(at least 1 red) Step 1: Total number of ways to draw 3 balls from 15 Total outcomes = C(15, 3) = 455 Step 2: Number of ways to get exactly 1 red We need 1 red and 2 non-red. – Choose 1 red from 4: C(4, 1) = 4 – Non-red balls = blue + green = 5 + 6 = 11 Choose 2 non-red from 11: C(11, 2) = 55 Favourable ways for exactly 1 red = 4 × 55 = 220 So, P(exactly 1 red) = 220 / 455 Step 3: Number of ways to get at least 1 red “At least 1 red” = Total ways − “no red”. “No red” ⇒ all 3 balls from non-red (11 balls): Ways with no red = C(11, 3) = 165 So, Ways with at least 1 red = Total − No red = 455 − 165 = 290 Therefore, P(at least 1 red) = 290 / 455 Step 4: Required conditional probability P(exactly 1 red | at least 1 red) = (220 / 455) ÷ (290 / 455) = 220 / 290 = 22 / 29