<p>An urn contains nine balls of which three are red, four are blue, and two are green. Three balls are drawn at random without replacement from the urn. The probability that the three balls have different colours is</p>
<p>(1) \(\dfrac{2}{23}\)</p>
<p>(2) \(\dfrac{1}{3}\)</p>
<p>(3) \(\dfrac{2}{7}\)</p>
<p>(4) \(\dfrac{1}{21}\)</p>
Step-by-Step Solution
Key Concept: For three balls of different colours (one red, one blue, one green), use combinations to count favorable outcomes: C(3,1)×C(4,1)×C(2,1), then divide by total ways to draw 3 balls from 9, which is C(9,3).
<p><strong>Step 1:</strong> Find total ways to draw 3 balls from 9 without replacement.</p><p>Total outcomes = C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 84</p><p><strong>Step 2:</strong> Count favorable outcomes (one ball of each colour).</p><p>We need: 1 red from 3, 1 blue from 4, 1 green from 2</p><p>Favorable outcomes = C(3,1) × C(4,1) × C(2,1) = 3 × 4 × 2 = 24</p><p><strong>Step 3:</strong> Calculate the probability.</p><p>P(different colours) = 24/84 = 2/7</p><p>∴ Answer: C</p>
Correct Answer: C