<p>A box contains 24 identical balls of which 12 are white and 12 are black. The balls are drawn at random from the box one at a time with replacement. The probability that a white ball is drawn for the 4th time on the 7th draw is</p>
Step-by-Step Solution
Key Concept: This is a negative binomial probability problem: we need exactly 3 white balls in the first 6 draws, then a white ball on the 7th draw. The probability of white on each draw is 1/2.
<p><strong>Step 1:</strong> Identify the constraint. The 4th white ball must appear on the 7th draw. This means exactly 3 white balls in the first 6 draws, and the 7th draw must be white.</p><p><strong>Step 2:</strong> Each ball is drawn with replacement, so P(white) = 12/24 = 1/2 and P(black) = 1/2 on each draw.</p><p><strong>Step 3:</strong> Probability of exactly 3 white balls in first 6 draws: $\binom{6}{3}\left(\frac{1}{2}\right)^3\left(\frac{1}{2}\right)^3 = \binom{6}{3}\left(\frac{1}{2}\right)^6$</p><p><strong>Step 4:</strong> Calculate: $\binom{6}{3} = \frac{6!}{3!3!} = 20$</p><p><strong>Step 5:</strong> Probability of white on 7th draw: $\frac{1}{2}$</p><p><strong>Step 6:</strong> Total probability: $20 \times \frac{1}{64} \times \frac{1}{2} = \frac{20}{128} = \frac{5}{32}$</p><p>∴ Answer: C</p>
Correct Answer: C