Question

20. Based on student records, 25% of the students at a large high school have a

GPA of 3.5 or better, 16% of the students are currently enrolled in at least one
AP class, and 12% of the students have a GPA of 3.5 or better and are enrolled
in at least one AP class. If we select one student at random, what is the
probability that the student has a GPA lower than 3.5 and is not taking any AP
classes?
(A) 0.29
(B) 0.47
(C) 0.53
(D) 0.63
(E) 0.71

Answer 1

(E) 0.71

Explanation:

P(not GPA > 3.5 and Not AP) = 1 – P(GPA > 3.5 or AP) = 1 – (0.25 + 0.16 – 0.12) = 0.71.

Answer 2

(E) 0.71

Explanation:

Let's call A the event that a student has GPA of 3.5 or better, A' the event that a student has GPA lower than 3.5, B the event that a student is enrolled in at least one AP class and B' the event that a student is not taking any AP class.

So, the probability that the student has a GPA lower than 3.5 and is not taking any AP classes is calculated as:

P(A'∩B') = 1 - P(A∪B)

it means that the students that have a GPA lower than 3.5 and are not taking any AP classes are the complement of the students that have a GPA of 3.5 of better or are enrolled in at least one AP class.

Therefore, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

Where the probability P(A) that a student has GPA of 3.5 or better is 0.25, the probability P(B) that a student is enrolled in at least one AP class is 0.16 and the probability P(A∩B) that a student has a GPA of 3.5 or better and is enrolled in at least one AP class is 0.12

So, P(A∪B) is equal to:

P(A∪B) = P(A) + P(B) - P(A∩B)

P(A∪B) = 0.25 + 0.16 - 0.12

P(A∪B) = 0.29

Finally, P(A'∩B') is equal to:

P(A'∩B') = 1 - P(A∪B)

P(A'∩B') = 1 - 0.29

P(A'∩B') = 0.71