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.

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

is there supposed to be a diagram or a graph for this one? ?

step-by-step explanation:

yes

step-by-step explanation:

so of course you want to start by plugging the 5 in which goes to 7(5)+3(5)=50 because 7*5=35 and 3*5=15 and then add them together, then plug in the 5 to the other equation as well 9(5)+5=50 because 9*5= 45 and then add 5 and so yes they are equivalent.