Early in August an undergraduate college discovers that it can accommodate a few extra students. Enrolling these additional students would provide a substantial increase in revenue without increasing the operating costs of the college; that is, no new classes would have to be added. From past experience, the college knows that 40% of those students will actually enroll. a) What is the probability that at most six students will enroll if the college offers admission to ten more students? b) What is the probability that more than 12 will actually enroll if admission is offered to 20 students? c) If 70% of those students admitted actually enroll, what is the probability that at least 12 out of 15 students will actually enroll?

A. P (x ≤ 6) = 0.9452

B. P (x > 12) = 0.021

C. P (x ≥ 12) = 0.2968

Step-by-step explanation:

Let's recall that in a binomial distribution the probability of success p in each trial is a fixed value and the result of each trial is independent of any previous trial.

A. Let's find out the probability that at most six students will enroll if the college offers admission to ten more students, using the following binomial distribution table:

Binomial distribution (n=10, p=0.4)

f (x) F (x) 1 - F (x)

x Pr[X = x] Pr[X ≤ x]

0 0.0060 0.0060

1 0.0403 0.0464

2 0.1209 0.1673

3 0.2150 0.3823

4 0.2508 0.6331

5 0.2007 0.8338

6 0.1115 0.9452

7 0.0425 0.9877

8 0.0106 0.9983

9 0.0016 0.9999

10 0.0001 1.0000

P (x ≤ 6) = 0.9452

B. Let's find that more than 12 will actually enroll if admission is offered to 20 students, using this second binomial distribution table:

Binomial distribution (n=20, p=0.4)

f (x) F (x) 1 - F (x)

x Pr[X = x] Pr[X ≤ x]

0 0.0000 0.0000

1 0.0005 0.0005

2 0.0031 0.0036

3 0.0123 0.0160

4 0.0350 0.0510

5 0.0746 0.1256

6 0.1244 0.2500

7 0.1659 0.4159

8 0.1797 0.5956

9 0.1597 0.7553

10 0.1171 0.8725

11 0.0710 0.9435

12 0.0355 0.9790

13 0.0146 0.9935

14 0.0049 0.9984

15 0.0013 0.9997

16 0.0003 1.0000

17 0.0000 1.0000

18 0.0000 1.0000

19 0.0000 1.0000

20 0.0000 1.0000

P (x ≤ 12) = 0.979, then:

P (x > 12) = 1 - 0.979 = 0.021

C. Let's find the probability that at least 12 out of 15 students will actually enroll if 70% of those students admitted actually enroll, using this new binomial distribution table:

Binomial distribution (n=15, p=0.7)

f (x) F (x) 1 - F (x)

x Pr[X = x] Pr[X ≤ x]

0 0.0000 0.0000

1 0.0000 0.0000

2 0.0000 0.0000

3 0.0001 0.0001

4 0.0006 0.0007

5 0.0030 0.0037

6 0.0116 0.0152

7 0.0348 0.0500

8 0.0811 0.1311

9 0.1472 0.2784

10 0.2061 0.4845

11 0.2186 0.7031

12 0.1700 0.8732

13 0.0916 0.9647

14 0.0305 0.9953

15 0.0047 1.0000

P (x ≤ 11) = 0.7031, then:

P (x ≥ 12) = 1 - 0.7031 = 0.2969, or,

P (x ≥ 12) = P (12) + P (13) + P (14) + P (15)

P (x ≥ 12) = 0.17 + 0.0916 + 0.0305 + 0.0047

P (x ≥ 12) = 0.2968