Answer:
40.1% probability that he will miss at least one of them
Step-by-step explanation:
For each target, there are only two possible outcomes. Either he hits it, or he does not. The probability of hitting a target is independent of other targets. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
0.95 probaiblity of hitting a target
This means that 
10 targets
This means that 
What is the probability that he will miss at least one of them?
Either he hits all the targets, or he misses at least one of them. The sum of the probabilities of these events is decimal 1. So

We want P(X < 10). So

In which

40.1% probability that he will miss at least one of them
8/20 is greater than 2/10. If you were to make the denominators the same, by mulipling 2x2 and 10x2, you would get 4/20, which is less than 8/20.
3/5 c 3 + 3 is equivalently
When writing the equation keep in mind that 7 + x is said as a term that is squared so it has to be squared while it's still together; (7 + x)².