Suppose that 60% of the students do homework regularly. it is also known that 80% of students who had been doing homework regularly, end up doing well in the course. only 20% of students who had not been doing homework regularly end up doing well in the course. given that a student did well in the course, what is the probability that the student had been doing homework regularly? 0.286 0.857 0.143 0.429

0.857

Step-by-step explanation:

Let H represents the event of doing homework regularly,

And, E represents the event of doing well in the course,

Then, According to the question,

60% of the students do homework regularly.

⇒ The probability of doing homework regularly, P(H) = 60% = 0.60

⇒The probability of not doing the homework regularly, P(H') = 1 - P(H) = 1 - 0.6 = 0.4

Also, 80% of students who had been doing homework regularly, end up doing well in the course.

⇒

20% of students who had not been doing homework regularly end up doing well in the course.

⇒  

⇒

Thus, By the definition of conditional probability,

The probability that the student had been doing homework regularly,given that a student did well in the course,

#15. square

step-by-step explanation:

because quadrilatural (idk how to spell it) means somthing that is not 3d or like, equal sides (or somthing like that   you know what i meanz)

hope this halps!

bai!

i believe the correct answer is b.