I have a following exercise:
At the exam there is $0.7$ probability that student has prepared and $0.3$ that he has not prepared. Those who are prepared have $0.9$ probability of success, those who have not prepared have $0.2$ probability of success. What is the probability that:
- randomly selected student will succeed;
- student who passed the exam has not prepared for it;
- student who did not pass the exam has prepared for it.
I think that solution to 1) is simply $0.7*0.9 + 0.3*0.2 = 0.69$. Chance of succeeding in each group summed - because the sets are disjunct.
I can't figure out 2 and 3 - can you please give me an advice? Thank you.
It will probably require using formula $P(A|B) = P(A \cap B) / P(B)$. In 2) I consider that A is event of passing the exam and B of not being prepared for it; $P(A|B) = P(A \cap B)/0.3$. But I don't know how to compute $P(A \cap B)$.