An anti-aircraft gun fires up to four shots at a receding plane. The hit probabilities on the 1st, 2nd, 3rd, and 4th shots are 0.4, 0.3, 0.2, and 0.1 respectively (independent). What is the probability the plane is hit at least once when all four shots are fired?

Introduction / Context:
When several independent attempts can cause success, the probability of at least one success is 1 minus the probability of zero successes. Here each shot has a different success probability.

Given Data / Assumptions:

• P(hit on 1st) = 0.4, P(hit on 2nd) = 0.3, P(hit on 3rd) = 0.2, P(hit on 4th) = 0.1.
• Shot results are independent.

Concept / Approach:
P(at least one hit) = 1 − Π P(miss on shot i). Miss probabilities are the complements: 0.6, 0.7, 0.8, 0.9.

Step-by-Step Calculation:
P(no hit) = 0.6 * 0.7 * 0.8 * 0.9 = 0.3024.Therefore, P(at least one hit) = 1 − 0.3024 = 0.6976.

Verification / Alternative check:
Any inclusion–exclusion computation must yield the same value; using the complement avoids lengthy term expansions.

Why Other Options Are Wrong:
Values like 0.6872 or 0.4379 arise from arithmetic slips or partial sums of inclusion–exclusion.

Common Pitfalls:
Adding hit probabilities directly (invalid due to overlaps) instead of using the complement of “no hit.”

Final Answer:
0.6976