Question

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?

Answer 1

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

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

• x is the number of successes.

• n is the number of trials.

• p is the probability of a success on a single trial.

In this problem:

• Hits the target 4 times out of 5, hence $p = \frac{4}{5} = 0.8$.

• He fires 4 shots, hence $n = 4$.

Item a:

This probability is:

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4)$

Then:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 2) = C_{4,2}.(0.8)^{2}.(0.2)^{2} = 0.1536$

$P(X = 3) = C_{4,3}.(0.8)^{3}.(0.2)^{1} = 0.4096$

$P(X = 4) = C_{4,4}.(0.8)^{4}.(0.2)^{0} = 0.4096$

$P(X \geq 2) = P(X = 2) + P(X = 3) + P(X = 4) = 0.1536 + 0.4096 + 0.4096 = 0.9728$

0.9728 = 97.28% probability of more than 2 hits.

Item b:

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

$P(X < 2) = 1 - P(X \geq 2) = 1 - 0.9728 = 0.0272$

0.0272 = 2.72% probability of at least 3 misses.

To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377