A marksman hits a target with probability 3/4 on each shot, independently. He fires 5 shots. What is the probability that he hits the target at least 3 times?

Introduction / Context:
We have a binomial model X ~ Binomial(n = 5, p = 3/4). The required probability is P(X ≥ 3) = P(3) + P(4) + P(5).

Given Data / Assumptions:

• n = 5 independent trials.
• p = 3/4 (success = hit).

Concept / Approach:
Use the binomial probability mass formula: P(X = k) = C(n, k) * p^k * (1 − p)^(n − k). Sum for k = 3, 4, 5.

Step-by-Step Solution:
P(3) = C(5, 3) * (3/4)^3 * (1/4)^2 = 10 * 27/64 * 1/16 = 270/1024.P(4) = C(5, 4) * (3/4)^4 * (1/4) = 5 * 81/256 * 1/4 = 405/1024.P(5) = C(5, 5) * (3/4)^5 = 1 * 243/1024 = 243/1024.Sum = (270 + 405 + 243)/1024 = 918/1024 = 459/512.

Verification / Alternative check:
Complement: 1 − [P(0) + P(1) + P(2)] yields the same 459/512 after computation.

Why Other Options Are Wrong:
The other fractions do not equal the computed exact value 459/512.

Common Pitfalls:
Arithmetic slips in powers or forgetting coefficients C(5, k). Keeping everything over 1024 reduces mistakes.

Final Answer:
459/512