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

Answer:

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.

Step-by-step explanation:

Conditional Probability

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

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

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)

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09}{0.72} = 0.125$

So

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