A (blindfolded) marksman finds that on the average he hits the target 4 times out of 5. If he fires 4 shots, what

is the probability of :

(a) more than 2 hits?

(b) at least 3 misses?

Using the binomial distribution, it is found that there is a:

a) 0.9728 = 97.28% probability of more than 2 hits.

b) 0.0272 = 2.72% probability of at least 3 misses.

For each shot, there are only two possible outcomes, either he hits the target, or he does not. The probability of hitting the target on a shot is independent of any other shot, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

The parameters are:

In this problem:

Item a:

This probability is:

Then:

0.9728 = 97.28% probability of more than 2 hits.

Item b:

At least 3 misses is less than 2 hits, hence:

0.0272 = 2.72% probability of at least 3 misses.

To learn more about the binomial distribution, you can take a look at link

answer: answer is c!

ok i got to organize the

3/15 and 12/55

3*

aha

3*4 equals 12 but not

a: no

b:

8/24 12/35

no

gautie

it's c

5/18 = 25/90

5*5=25

18*5 = 90

hope that you!

one more happy customer!

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step-by-step explanation: