Probability
Independent Events
Grade 12
Question:
<p>A soldier is firing at a moving target. He fires four shots. The probability of hitting the target at the first, second, third and fourth shots are 0.6, 0.4, 0.2 and 0.1 respectively. What is the probability that he hits the target?</p>
<p>(a) \(\frac{517}{625}\)</p>
<p>(b) \(\frac{3}{625}\)</p>
<p>(c) \(\frac{105}{625}\)</p>
<p>(d) none of these</p>
Step-by-Step Solution
Key Concept: Use the complement rule: P(at least one hit) = 1 - P(no hits). Since shots are independent, P(no hits) = product of individual miss probabilities.
<p><strong>Step 1:</strong> Identify what we need. 'He hits the target' means <strong>at least one hit</strong> in four shots.</p><p><strong>Step 2:</strong> Use complement rule: P(at least one hit) = 1 - P(no hits)</p><p><strong>Step 3:</strong> Find P(no hits) for each shot (miss probabilities):</p><ul><li>P(miss shot 1) = 1 - 0.6 = 0.4</li><li>P(miss shot 2) = 1 - 0.4 = 0.6</li><li>P(miss shot 3) = 1 - 0.2 = 0.8</li><li>P(miss shot 4) = 1 - 0.1 = 0.9</li></ul><p><strong>Step 4:</strong> Since shots are independent:</p><p>P(all 4 misses) = 0.4 × 0.6 × 0.8 × 0.9 = 0.1728</p><p><strong>Step 5:</strong> Apply complement:</p><p>P(at least one hit) = 1 - 0.1728 = <strong>0.8272</strong></p><p>∴ Answer: D</p>
Correct Answer: D