Probability of at least one

The probability that a bomber hits its target on any particular mission is .80. Four bombers are sent after the same target.

What is the probability:

a) They all hit the target? Ans: (0.8) ^ 4 = 0.4096
b) None hit the target? Ans: (0.2) ^ 4 = 0.0016
c) At least one hits the target? Ans: 1 – 0.0016 = 0.9984 ( I don’t get this one)

I know this is supposedly a simple question, but I really can't quite figure out why the working is such for c. Any pointer will be most helpful.

Thanks!

For part c, the opposite of 'at least one' is none hit the target.

Figure out the probability that none hit the target and subtract from 1.

\(\displaystyle (0.20)^{4}=0.0016\). See?.

Thanks Galactus

Appreciate your input!