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 $P(A \cap B)$.
Best Answer
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 $\frac{60}{690}$.