Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho

Answer:

a) The likelihood that a student will consistently complete homework and successfully pass the course = P(H n P) = 0.57

b) The likelihood that a student will not consistently complete homework nor will pass the course = P(H' n P') = 0.12

c) The two events, passing the course and regularly completing homework, are not mutually exclusive. Refer to the explanation for details.

d) The two events, passing the course and regularly completing homework, are not independent. Refer to the explanation for details.

Step-by-step explanation:

Let H be the event of consistently doing homework.

Let P be the event of passing the course.

- 60% of the students do homework consistently

P(H) = 60% = 0.60

- 95% of students who regularly do homework tend to pass the course

P(P|H) = 95% = 0.95

- Furthermore, it is known that 85% of students pass the course.

P(P) = 85% = 0.85

a) The probability a student will do homework regularly and also pass the course is given as P(H n P)

The conditional probability of A happening given that B has occurred, P(A|B), is expressed as

P(A|B) = P(A n B) ÷ P(B)

Thus, this implies

P(A n B) = P(A|B) × P(B)

Thus,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will not do homework regularly nor pass the course = P(H' n P')

From Set Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Having P(H n P) = 0.57 (from part a),

Note also that

P(H) = P(H n P') + P(H n P) (as events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

Thus, P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (considering events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

Hence, P(H' n P) = 0.85 - 0.57 = 0.28

Therefore, P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1 becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

So, P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Justify your answer.

Two events are deemed mutually exclusive when they cannot occur simultaneously. The mathematical notation indicating two events A and B as mutually exclusive is that if A and B are mutually exclusive then

P(A n B) = 0.

However, P(H n P) has been computed to be 0.57, thus P(H n P) = 0.57 ≠ 0.

Therefore, these events are not mutually exclusive.

Two events are labeled independent if the probability of one does not impact the occurrence of the other. It is mathematically determined that two events A and B are independent when

P(A|B) = P(A) and P(B|A) = P(B)

P(A n B) = P(A) × P(B)

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

Hope this helps!!!