Probability of at least one

[dotts05]

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!

 

[galactus]

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?.

 

[dotts05]

Thanks Galactus

Appreciate your input!