Probability
Multinomial Distribution
Grade 12
Question:
<p>10 balls are thrown into three boxes namely Box 1, Box 2, and Box 3 with respective probabilities 1/4, 1/4 and 1/2. What is the probability that out of 10 balls, 1\(^{\text{st}}\) box will get 2, 2\(^{\text{nd}}\) will get 3 and 3\(^{\text{rd}}\) box will get 5 balls?</p>
<p>(a) \(\dfrac{10!}{(2!)(3!)(5!)(2^{15})}\)</p>
<p>(b) \(\dfrac{1}{2^{15}}\)</p>
<p>(c) \(\dfrac{1}{2^{14}}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: This is a multinomial probability problem where we need to find the probability of a specific distribution of 10 independent trials across 3 categories with given probabilities. Use the multinomial formula: P = (n!)/(n₁!n₂!n₃!) × p₁^n₁ × p₂^n₂ × p₃^n₃
<p><strong>Step 1:</strong> Identify the parameters:</p><ul><li>Total balls: n = 10</li><li>Box 1: n₁ = 2 balls, probability p₁ = 1/4</li><li>Box 2: n₂ = 3 balls, probability p₂ = 1/4</li><li>Box 3: n₃ = 5 balls, probability p₃ = 1/2</li></ul><p><strong>Step 2:</strong> Apply the multinomial probability formula:</p><p>P(2,3,5) = [10!/(2!×3!×5!)] × (1/4)² × (1/4)³ × (1/2)⁵</p><p><strong>Step 3:</strong> Calculate the multinomial coefficient:</p><p>10!/(2!×3!×5!) = 3,628,800/(2×6×120) = 3,628,800/1,440 = 2,520</p><p><strong>Step 4:</strong> Calculate the probability terms:</p><p>(1/4)² = 1/16</p><p>(1/4)³ = 1/64</p><p>(1/2)⁵ = 1/32</p><p><strong>Step 5:</strong> Multiply all components:</p><p>P = 2,520 × (1/16) × (1/64) × (1/32) = 2,520/(16×64×32) = 2,520/32,768 = 315/4,096</p><p>∴ Answer: A</p>
Correct Answer: A