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

2 x^2 -3x+p

sum of roots = 3/2

let other root be x

3+ x = 3/2

x= 3/2 - 3 = -3/2

product of roots = p/2

3× -3/2 = p/2

-9 = p

⭐⭐

nada

step-by-step explanation: