# Conditional probability exercise

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:

1. randomly selected student will succeed;
2. student who passed the exam has not prepared for it;
3. 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 $P(A \cap B)$.

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Note that for question 1. you computed $P(A)$ by summing $P(A) = P(A \cap B) + P(A \cap \neg B) = 0.3 \cdot 0.2 + 0.7 \cdot 0.9$. So you already computed $P(A \cap B)$... – TMM Jan 5 '13 at 17:20

We do 2. A similar procedure handles 3.

We want the probability that a student who passes did not prepare. So we want the probability she did not prepare, given that she passed.

Let $P$ be the event "she passes," and let $N$ be the event "she did not prepare." In symbols, we want $\Pr(N|P)$. This is the key observation, the rest is downhill.

By the formula that you quoted, we have $$\Pr(N|P)=\frac{\Pr(N\cap P)}{\Pr(P)}.$$ You found $\Pr(P)$ correctly in part 1 of the problem.

To find $\Pr(N\cap P)$, note that the probability of $N$ is $0.3$. Given that she does not prepare, the probability the passes is $0.2$. So the probability she does not prepare and passes is $(0.3)(0.2)$.

An informal version: We have a group of $1000$ students. About $700$ prepared, and of these about $630$ passed. About $300$ did not prepare, and about $60$ of them passed. So about $690$ students passed. If we confine attention (restrict the sample space) to the $690$ who passed, about $60$ did not prepare. So the probability a student who passed did not prepare is $\frac{60}{690}$.

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Thank you, I assumed I am looking for Pr(P|N) which was the wrong way. – ondrejsl Jan 5 '13 at 18:03
@OndrejSlamecka: I had a mistake in the informal part. About $690$ passed. Changed numbers suitably. – André Nicolas Jan 5 '13 at 18:06
@OndrejSlamecka: Yes, deciding what one needs to compute is, as I mentioned, the key step. We are confining attention to students who passed. – André Nicolas Jan 5 '13 at 18:08