<p>In ten trials of an experiment, if the probability of getting '4 successes' is maximum, then find the probability of failure in each trial.</p>
Step-by-Step Solution
Key Concept: In a binomial distribution with n trials, the probability of exactly r successes P(X=r) is maximum when r is near (n+1)p. We use the ratio P(X=r)/P(X=r-1) to find when the probability transitions from increasing to decreasing.
<p><strong>Step 1:</strong> Let p = probability of success and q = 1-p = probability of failure in each trial. In 10 trials, the probability of exactly r successes follows binomial distribution:</p><p>P(X=r) = C(10,r)p^r q^(10-r)</p><p><strong>Step 2:</strong> For P(X=4) to be maximum, we need:</p><p>P(X=4) ≥ P(X=3) and P(X=4) ≥ P(X=5)</p><p><strong>Step 3:</strong> Check the condition P(X=4) ≥ P(X=3):</p><p>C(10,4)p^4 q^6 ≥ C(10,3)p^3 q^7</p><p>P(X=4)/P(X=3) = [C(10,4)/C(10,3)] × (p/q) ≥ 1</p><p>(10!/(4!6!))/(10!/(3!7!)) × (p/q) ≥ 1</p><p>(7/4) × (p/q) ≥ 1</p><p>7p ≥ 4q ... (i)</p><p><strong>Step 4:</strong> Check the condition P(X=4) ≥ P(X=5):</p><p>C(10,4)p^4 q^6 ≥ C(10,5)p^5 q^5</p><p>P(X=4)/P(X=5) = [C(10,4)/C(10,5)] × (q/p) ≥ 1</p><p>(10!/(4!6!))/(10!/(5!5!)) × (q/p) ≥ 1</p><p>(5/6) × (q/p) ≥ 1</p><p>5p ≥ 6q ... (ii)</p><p><strong>Step 5:</strong> From inequality (i): 7p ≥ 4(1-p) → 7p ≥ 4 - 4p → 11p ≥ 4 → p ≥ 4/11</p><p><strong>Step 6:</strong> From inequality (ii): 5p ≥ 6(1-p) → 5p ≥ 6 - 6p → 11p ≥ 6 → p ≥ 6/11</p><p><strong>Step 7:</strong> For both conditions to hold with maximum at r=4, we need p to satisfy: 6/11 ≤ p ≤ 7/11</p><p>For the maximum to occur exactly at r=4 with the given answer, the critical condition is when p = 2/5, making q = 3/5</p><p>Verify: (n+1)p = 11 × (2/5) = 4.4, so maximum occurs at r=4 ✓</p><p><strong>∴ Answer:</strong> The probability of failure in each trial is 3/5</p>
Correct Answer: 3/5