Question

In Prof. Leahy's class, if a student does homework most days, the chance of passing the course is 90%. On the other hand, if a student does not do homework most days, the chance of passing the course is only 30%. Prof. Leahy claims that 70% of his students do homework on a regular basis. If a student is chosen at random from Prof. Leahy's class, find the following probabilities:

a. p(the student did not do homework and he/she passed the course)
b. p(the student did not do homework given that she/he passed the course)

Answer 1

a) p(the student did not do homework and he/she passed the course) = 0.09

b) p(the student did not do homework given that she/he passed the course) = 0.125.

Explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

`P(B|A) = (P(A \cap B))/(P(A))`

In which

P(B|A) is the probability of event B happening, given that A happened.

`P(A \cap B)` is the probability of both A and B happening.

P(A) is the probability of A happening.

a. p(the student did not do homework and he/she passed the course)

If a student does not do homework most days, the chance of passing the course is only 30%. 100 - 70 = 30% don't do homework on a regular basis.

So

0.3*0.3 = 0.09

p(the student did not do homework and he/she passed the course) = 0.09

b. p(the student did not do homework given that she/he passed the course)

Conditional probability.

Event A: Passed the course

Event B: Did not do homework.

p(the student did not do homework and he/she passed the course) = 0.09

This means that
`P(A \cap B) = 0.09`

Probability that the student passes the course:

90% of 70%(do homework)

30% of 30%(do not do homework).

This means that:

`P(A) = 0.9*0.7 + 0.3*0.3 = 0.72`

p(the student did not do homework given that she/he passed the course)

`P(B|A) = (P(A \cap B))/(P(A)) = (0.09)/(0.72) = 0.125`

So

p(the student did not do homework given that she/he passed the course) = 0.125.