Samir is an expert marksman. When he takes aim at a particular target on the shooting range, there is a 0.950.950, point, 95 probability that he will hit it. One day, Samir decides to attempt to hit 101010 such targets in a row.
Assuming that Samir is equally likely to hit each of the 101010 targets, what is the probability that he will miss at least one of them?
Round your answer to the nearest tenth.

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.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

0.95 probaiblity of hitting a target

This means that [tex]p = 0.95[/tex]

10 targets

This means that [tex]n = 10[/tex]

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

[tex]P(X = 10) + P(X < 10) = 1[/tex]

We want P(X < 10). So

[tex]P(X < 10) = 1 - P(X = 10)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 10) = C_{10,10}.(0.95)^{10}.(0.05)^{0} = 0.5987[/tex]

[tex]P(X < 10) = 1 - P(X = 10) = 1 - 0.5987 = 0.401[/tex]

40.1% probability that he will miss at least one of them