Marksman with hit probability 1/5 per shot and 10 shots. What is the probability of at least one hit?

Introduction / Context:This is the same “at least one success” structure with p = 1/5 (hit) and q = 4/5 (miss), repeated over 10 independent trials.

Given Data / Assumptions:

• Hit probability p = 1/5; miss probability q = 4/5.
• Number of shots n = 10; independence assumed.

Concept / Approach:Complement rule: P(≥1 hit) = 1 − q^n = 1 − (4/5)^{10}.

Step-by-Step Solution:Compute (4/5)^{10} if a numeric value is needed; algebraic form is acceptable.

Verification / Alternative check:As n increases, (4/5)^n decreases, so the at-least-one probability increases, consistent with intuition.

Why Other Options Are Wrong:(4/5)10 is missing an exponent; 1/(5)10 is not a probability expression; 1 − 1/(5)10 is unrelated; (1/5)10 would be all-hits if it had an exponent.

Common Pitfalls:Dropping exponents or misreading (4/5)^10 as 4/(5×10).

Final Answer:1 - (4/5)^10