Probability
Classical Probability
Grade 12
Question:
<p>A box contains 2 black, 4 white, and 3 red balls. One ball is drawn at random from the box and kept aside. From the remaining balls in the box, another ball is drawn at random and kept aside the first. This process is repeated till all the balls are drawn from the box. The probability that the balls drawn are in the sequence of 2 black, 4 white, and 3 red is</p>
<p>1/1260</p>
<p>1/7560</p>
<p>1/126</p>
<p>none of these</p>
Step-by-Step Solution
Key Concept: The probability of drawing balls in a specific sequence without replacement equals the product of conditional probabilities at each step: (2/9) × (1/8) × (4/7) × (3/6) × ... The key insight is recognizing this equals 2!×4!×3!/9!, which simplifies to the reciprocal of the multinomial coefficient.
<p><strong>Step 1:</strong> Total balls = 2 + 4 + 3 = 9. We need all 2 black balls first, then all 4 white, then all 3 red.</p><p><strong>Step 2:</strong> Probability of drawing 2 black balls first (in sequence):</p><p>P(1st black) × P(2nd black | 1st black) = (2/9) × (1/8)</p><p><strong>Step 3:</strong> Probability of drawing 4 white balls next (in sequence):</p><p>P(1st white) × P(2nd white) × P(3rd white) × P(4th white) = (4/7) × (3/6) × (2/5) × (1/4)</p><p><strong>Step 4:</strong> Probability of drawing 3 red balls last (in sequence):</p><p>P(1st red) × P(2nd red) × P(3rd red) = (3/3) × (2/2) × (1/1) = 1</p><p><strong>Step 5:</strong> Total probability = (2/9) × (1/8) × (4/7) × (3/6) × (2/5) × (1/4) × 1</p><p><strong>Step 6:</strong> This equals (2! × 4! × 3!)/(9!) = (2 × 24 × 6)/362880 = 288/362880 = 1/1260</p><p>∴ Answer: A</p>
Correct Answer: A