<p>In a sequence of independent trials, the probability of success in one trial is \(\dfrac{1}{4}\). Find the probability that the second success takes place on or after the fourth trial.</p>
Step-by-Step Solution
Key Concept: The second success occurs on or after the fourth trial means the first three trials must contain at most one success. Use the complement: P(second success ≥ 4th trial) = P(≤1 success in first 3 trials).
<p><strong>Step 1:</strong> Interpret the condition. The second success occurs on or after the 4th trial if and only if there is at most 1 success in the first 3 trials.</p><p><strong>Step 2:</strong> Calculate P(at most 1 success in 3 trials) where p = 1/4, q = 3/4.</p><p>P(0 successes) = C(3,0)(1/4)⁰(3/4)³ = 1 · 1 · 27/64 = 27/64</p><p>P(1 success) = C(3,1)(1/4)¹(3/4)² = 3 · (1/4) · (9/16) = 27/64</p><p><strong>Step 3:</strong> Add the probabilities.</p><p>P(second success ≥ 4th trial) = P(0 successes) + P(1 success) = 27/64 + 27/64 = 54/64 = <strong>27/32</strong></p><p><strong>Alternative approach:</strong> Using negative binomial: P(exactly 1 success in first 3 trials AND 2nd success later) + P(0 successes in first 3 trials) gives the same result.</p><p>∴ Answer: <strong>27/32</strong></p>
Correct Answer: 27