<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 colour is</p>
Step-by-Step Solution
Key Concept: When drawing without replacement, calculate the probability of getting one ball of each color by considering all possible orderings. The total ways to select 3 balls from 9 is C(9,3), and favorable outcomes are the product of individual selections multiplied by the number of arrangements (3!).
<p><strong>Step 1:</strong> Find total ways to draw 3 balls from 9 without replacement.</p><p>Total ways = C(9,3) = 9!/(3!×6!) = (9×8×7)/(3×2×1) = 84</p><p><strong>Step 2:</strong> Find favorable outcomes (one red, one blue, one green).</p><p>Ways to select 1 red from 3: C(3,1) = 3</p><p>Ways to select 1 blue from 4: C(4,1) = 4</p><p>Ways to select 1 green from 2: C(2,1) = 2</p><p>Favorable outcomes = 3 × 4 × 2 = 24</p><p><strong>Step 3:</strong> Calculate probability.</p><p>P(different colors) = 24/84 = 2/7</p><p>∴ Answer: A (2/7)</p>
Correct Answer: A