Question

For a binomial distribution with a sample size equal to 10 and a probability of a success equal to​ 0.30, what is the probability that the sample will contain exactly three​ successes? Use the binomial formula to determine the probability. Round to four decimal places.

Answer 1

Answer:

The probability that the sample will contain exactly three​ successes is 0.2668.

Step-by-step explanation:

Given information:

Sample size = 10

Probability of a success, p=0.30

Probability of a failure q=1-p = 1-0.30 = 0.70

The binomial formula to determine the probability is

$P(X=r)=^nC_rp^rq^{n-r}$

where, n is the sample size, r is required number of success, p is probability of success and q is probability of failure.

We need to find the probability that the sample will contain exactly three​ successes.

$P(X=3)=^{10}C_3(0.30)^3(0.70)^{10-3}$

$P(X=3)=^{10}C_3(0.30)^3(0.70)^{7}$

$P(X=3)=(120)(0.0022235661)$

$P(X=3)=0.266827932$

$P(X=3)\approx 0.2668$

Therefore the probability that the sample will contain exactly three​ successes is 0.2668.

Answer 2

Using the binomial distribution, it is found that there is a 0.2668 = 26.68% probability that the sample will contain exactly three​ successes.

What is the binomial distribution formula?

The formula is:

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

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

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:

• The sample size is of n = 10.

• The probability of a success is of p = 0.3.

The probability that the sample will contain exactly three​ successes is P(X = 3), hence:

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

$P(X = x) = C_{10,3}.(0.3)^{3}.(0.7)^{7} = 0.2668$

More can be learned about the binomial distribution at https://brainly.com/question/24863377