Answer:
a.) 0.57
b.) 0.12
Step-by-step explanation:
Let the event that students do homework regularly be the event A
It is given that the probability that students do homework regularly , p(A)= 0.6
Let the event that students pass the course be the event B.
It is given that the probability that students pass the course, p(B) = 0.85
It is also given that students who do their homework also pass the course with a probability of 0.95.
Therefore p( B | A) =0.95
a.) The probability that a student will do the homework regularly and also pass the course
[tex]p( B | A) = \frac{p(A \cap B }{p(A)}[/tex]
∴ [tex]p(A \cap B) =[/tex] p(B | A) × p(A) = 0.95 × 0.6 = 0.57
b.) [tex]p((A\cup B)') = 1 - p(A \cup B) = 1 -\{ (p(A) + p(B) - p(A \cap B) \}\\[/tex]
Therefore [tex]p((A\cup B)') = 1 - \{ 0.6 + 0.85 - 0.57 \} = 1 - 0.88 = 0.12[/tex]
