Question

The admissions office of a private university released the following data for the preceding academic year: From a pool of 4200 male applicants, 30% were accepted by the university, and 30% of these subsequently enrolled. Additionally, from a pool of 3300 female applicants, 35% were accepted by the university, and 30% of these subsequently enrolled. What is the probability of each of the following?
a) A male applicant will be accepted by and subsequently will enroll in the university?

b) A student who applies for admissions will be accepted by the university?

c) A student who applies for admission will be accepted by the university and subsequently will enroll?

Answer 1

Answer:

(a) 0.09 (b) 0.322 (c) 0.0966

Step-by-step explanation:

Let's define first the following events

M: an applicant is a male

F: an applicant is a female

A: an applicant is accepted

E: an applicant is enrolled

S: the sample space

Now, we have a total of 7500 applicants, and from these applicants 4200 were male and 3300 were female. So,

P(M) = 0.56 and P(F) = 0.44, besides

P(A | M) = 0.3, P(E | A∩M) = 0.3, P(A | F) = 0.35, P(E| A∩F) = 0.3

(a) 0.09 = (0.3)(0.3) = P(A|M)P(E|A∩M)=P(E∩A∩M)/P(M)=P(E∩A | M)

(b) P(A) = P(A∩S) = P(A∩(M∪F))=P(A∩M)+P(A∩F)=P(A|M)P(M)+P(A|F)P(F)=(0.3)(0.56)+(0.35)(0.44)=0.322

(c) P(A∩E)=P(A∩E∩S)=P(A∩E∩(M∪F))=P(A∩E∩M)+P(A∩E∩F)=0.0504+P(E|A∩F)P(A|F)P(F)=0.0504+(0.3)(0.35)(0.44)=0.0966