Probability
Repeated Independent Trials
Grade 12
Question:
<p>If the probability of hitting a target by a shooter, in any shot, is \(1/3\), then the minimum number of independent shots at the target required by him so that the probability of hitting the target at least once is greater than \(5/6\), is ______.</p>
Step-by-Step Solution
Key Concept: Use the complement rule: P(at least one hit) = 1 - P(no hits). Set up the inequality 1 - (2/3)^n > 5/6 and solve for minimum n.
<p><strong>Step 1:</strong> Let n = number of shots required.</p><p>Probability of hitting in one shot = 1/3, so probability of missing in one shot = 2/3.</p><p><strong>Step 2:</strong> For n independent shots, probability of missing all n shots = (2/3)^n.</p><p><strong>Step 3:</strong> Probability of hitting at least once = 1 - (2/3)^n.</p><p><strong>Step 4:</strong> We need: 1 - (2/3)^n > 5/6</p><p><strong>Step 5:</strong> Rearranging: (2/3)^n < 1 - 5/6 = 1/6</p><p><strong>Step 6:</strong> Taking natural log: n·ln(2/3) < ln(1/6)</p><p>Since ln(2/3) is negative: n > ln(1/6)/ln(2/3) = ln(6)/ln(3/2)</p><p><strong>Step 7:</strong> Calculating: n > ln(6)/ln(1.5) ≈ 1.7918/0.4055 ≈ 4.42</p><p><strong>Step 8:</strong> Since n must be an integer and n > 4.42, minimum n = 5.</p><p><strong>Verification:</strong> For n=5: 1-(2/3)^5 = 1-32/243 = 211/243 ≈ 0.868 > 5/6 ≈ 0.833 ✓</p><p><strong>For n=4:</strong> 1-(2/3)^4 = 1-16/81 = 65/81 ≈ 0.802 < 5/6 ✗</p><p>∴ <strong>Answer: 5</strong></p>
Correct Answer: 5