A box contains 4 red balls, 5 blue balls and 6 green

Solution

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