Answer:
Step-by-step explanation:
Hello!
For a recent admissions class, an Ivy League college received 2851 applications for early admission. Of this group, it admitted 1033 students early, rejected 854 outright, and deferred 964 to the regular admission pool for further consideration.
In the past, this school has admitted 18% of the deferred early admission applicants during the regular admission process. Counting the students admitted early and the students admitted during the regular admission process, the total class size was 2375.
If the events:
E: the student applied for early admission and got it.
R: the student applied to the early admission and got outright rejected.
D: the student applied to the early admission and got deferred to regular admissions.
Total of early applications 2851
Early admitted 1033
Rejected 854
Deferred 964
Historically 18% of the deferred students get admitted.
a. To calculate the probabilities of each event, you have to divide the number of applications that correspond to the event by the total of applications received:
P(E)= 1033/2851= 0.36
P(R)= 854/2851= 0.30
P(D)= 964/2851= 0.34
b. Two events are mutually exclusive when the occurrence of one prevents the occurrence of the other in a single repetition of the experiment. This means that if we randomly choose one early application, the student can be "accepted early" or "deferred to regular admissions" but not the two of them at the same time, which means that E and D are mutually exclusive events.
Since both events are mutually exclusive and cannot happen at the same time, P(E∩D)= 0
c. In this case, the total "population" is not the number of early admissions received by the college you have to work using the total "population" of admitted students.
A: the admitted student was accepted during the early application.
P(A)= 1033/2375= 0.43
d. You need to calculate the probability of the student being admitted early or deferred and later admitted. This probability is a union of two events where "or" is symbolized with ∪.
P(E)= 0.36
Let the event B: the student was deferred and later admitted during the regular admission process. According to the text in the past 18% of the deferred students get admitted. So the probability of b is:
P(B)= 0.18
These two events are mutually exclusive, if the student was admitted early then he cannot be admitted during the regular admissions process.
P(E∪B) = P(E) + P(B)= 0.36 + 0.18= 0.54
I hope it helps!
