Question

At a local community college, 57% of students who enter the college as freshmen go on to graduate. Five freshmen are randomly selected.

a. What is the probability that none of them graduates from the local community college? (Do not round intermediate calculations Round your final answer to 4 decimal places Probability
b. What is the probability that at most four will graduate from the local community college? (Do not round intermediate calculations. Round your final answer to 4 decimal places.)
c. What is the expected number that will graduate? (Round your final answer to 2 decimal places)

Answer 1

To solve this problem, we can use the binomial probability formula. For part a, the probability that none of the five freshmen graduate is 0.43^5. For part b, we can calculate the probability that at most four of the freshmen graduate by adding the probabilities of exactly 0, 1, 2, 3, and 4 freshmen graduating. For part c, the expected number of freshmen that will graduate is calculated using the formula: n * p.

Step-by-step explanation:

To solve this problem, we can use the binomial probability formula:

P(X = k) = C(n, k) * pk * (1-p)(n-k)

where:

P(X = k) is the probability of getting exactly k successes in n trials

C(n, k) is the number of combinations of n items taken k at a time

p is the probability of success in a single trial

For part a, we want the probability that none of the five freshmen graduates. Since the probability of a single freshman graduating is 57%, the probability of a single freshman not graduating is 43%. So, the probability that none of the five freshmen graduates is:

P(X = 0) = C(5, 0) * (0.43)0 * (0.57)5 = 0.435

For part b, we want the probability that at most four of the freshmen graduate. We can calculate this by finding the probabilities of exactly 0, 1, 2, 3, and 4 freshmen graduating, and then adding them together:

P(X ≤ 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)

For part c, the expected number of freshmen that will graduate is given by the formula:

Expected value (E) = n * p = 5 * 0.57

Answer 2

a) 0.0147 = 1.47% probability that none of them graduates from the local community college.

b) 0.9398 = 93.98% probability that at most four will graduate from the local community college.

c) The expected number that will graduate is 2.85.

Step-by-step explanation:

For each student, there are only two possible outcomes. Either they will graduate, or they will not. The probability of a student graduating is independent of any other student graduating, which means that the binomial probability distribution is used 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.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

57% of students who enter the college as freshmen go on to graduate.

This means that
`p = 0.57`

Five freshmen are randomly selected.

This means that
`n = 5`

a. What is the probability that none of them graduates from the local community college?

This is P(X = 0). So

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(5,0).(0.57)^(0).(0.43)^(5) = 0.0147`

0.0147 = 1.47% probability that none of them graduates from the local community college.

b. What is the probability that at most four will graduate from the local community college?

This is:

`P(X \leq 4) = 1 - P(X = 5)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 5) = C_(5,5).(0.57)^(5).(0.43)^(0) = 0.0602`

So

`P(X \leq 4) = 1 - P(X = 5) = 1 - 0.0602 = 0.9398`

0.9398 = 93.98% probability that at most four will graduate from the local community college.

c. What is the expected number that will graduate?

The expected value of the binomial distribution is:

`E(X) = np`

In this question:

`E(X) = 5*0.57 = 2.85`

The expected number that will graduate is 2.85.