Question

In a certain large college course, past records show that grades of A, B, C, D, and F (which are the only grades assigned) are equally likely. If one student is chosen at random, what is:

(a) Pr(C)

(b) Pr(A or B)

(c) Pr(a grade better than D)

(d) Pr(A, B, C, D, or F)

(e) Pr(B and D)

(f) Pr(E)

If two students who do not know one another take the course described above, what are the following probabilities:

(a) Pr(two As)

(b) Pr(same grade)

(c) Pr(different grades)

(d) Pr(both better than D)

(e) Pr(both fail) = Pr(two Fs)

(f) Pr(one passes and one fails)

Answer 1

1.
Grades A, B, C, D, F are equally likely
P(A) = P(B) = P(C) = P(D) = P(F) = 0.2
(a)
Randomly selected student is assigned a grade C : P(C) = 0.2

(b)
P(A or B) = P(A) + P(B) - P(A and B)
P(A and B) = 0 , Since the events of assigning different grades are mutually exclusive.
P(A or B) = P(A) + P(B) = 0.2 + 0.2 = 0.4

(c)
P(a grade better than D) = P(A) + P(B) + P(C) = 0.2 + 0.2 + 0.2 = 0.6

(d)
P(A, B, C, D or F) = 1 (Mutually exclusive and exhaustive)

(e)
P(B and D) = 0 (Mutually Exclusive - One student cannot be assigned two grades)

(f)
P(E) = 0 (Only grades assigned are A, B, C, D and F)

2.
(a)
P(two As) = P(Both the students get A) = P(A)² = 0.2² = 0.04

(b)
P(same grade) = P(Both the students get the same grade)
Select a grade in ways and assign it to both the students.
P(same grade) = x 0.2² = 5 x 0.04 = 0.2

(c)
P(different grades) = 1 - P(same grades) = 1 - 0.2 = 0.8

(d)
P(both better than D) = [P(a grade better than D)] ²
P(a grade better than D) = 0.6 (from part 1.c)
P(both better than D) = 0.6² = 0.36

(e)
P(Both Fail) = P(two Fs) = 0.2² = 0.4

(f)
P(one passes and one fails)
P(pass) = P(A, B, C or D) = 0.8
P(fail) = P(D) = 0.2
P(one passes and one fails) = 2 x (0.8) x (0.2) = 0.32