<p>A bag contains 3 red, 4 black and 2 white balls. Three balls are drawn at random. What is the probability that the three balls have different colours?</p>
<p>(1) \(\dfrac{2}{7}\)</p>
<p>(2) \(\dfrac{1}{7}\)</p>
<p>(3) \(\dfrac{3}{7}\)</p>
<p>(4) \(\dfrac{4}{7}\)</p>
Step-by-Step Solution
Key Concept: For three balls of different colors, you must select exactly one red from 3, one black from 4, and one white from 2. Use the multiplication principle: favorable outcomes = C(3,1) × C(4,1) × C(2,1), and total outcomes = C(9,3).
<p><strong>Step 1:</strong> Find total number of balls: 3 red + 4 black + 2 white = 9 balls</p><p><strong>Step 2:</strong> Total ways to draw 3 balls from 9: C(9,3) = 9!/(3!×6!) = 84</p><p><strong>Step 3:</strong> For three balls of different colors, we need exactly one ball from each color:</p><ul><li>Ways to select 1 red from 3: C(3,1) = 3</li><li>Ways to select 1 black from 4: C(4,1) = 4</li><li>Ways to select 1 white from 2: C(2,1) = 2</li></ul><p><strong>Step 4:</strong> Favorable outcomes = C(3,1) × C(4,1) × C(2,1) = 3 × 4 × 2 = 24</p><p><strong>Step 5:</strong> Probability = 24/84 = 2/7</p><p>∴ Answer: A (or 2/7 depending on options)</p>
Correct Answer: A