The probability of a man hitting a target is 1/5. If the man fires 7 times, then what is the probability that he hits the target at least twice?

Using binomial distribution with n=7, p=1/5, q=4/5. P(at least 2 hits) = 1 - P(0 hits) - P(1 hit). P(0 hits) = (4/5)^7. P(1 hit) = 7·(1/5)·(4/5)^6 = (7/5)(4/5)^6. So P(at least 2) = 1 - (4/5)^7 - (7/5)(4/5)^6 = 1 - (4/5)^6[(4/5) + 7/5] = 1 - (4/5)^6·(11/5). Wait, that gives option (c). Let me recompute: (4/5)^7 = (4/5)(4/5)^6. So 1 - (4/5)(4/5)^6 - (7/5)(4/5)^6 = 1 - (4/5)^6[(4/5) + (7/5)] = 1 - (4/5)^6·(11/5) = 1 - (11/5)(4/5)^6. This matches option (c).