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4 May, 06:28

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

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Answers (2)
  1. 4 May, 06:41
    0
    (E) 0.71

    Step-by-step 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
  2. 4 May, 06:44
    0
    (E) 0.71

    Step-by-step 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.
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