# Conditional Probability question

• Juanriq

## Homework Statement

Students who actively participate in class are 4 times more likely to get a good grade than those who don't. 15% of students actively participate in class; 20% of students get a good grade. A person did not participate in class, what is the probability that he got a good grade?

## The Attempt at a Solution

Well, I define P(G): student got a good grade; P(P): student participates. Obviously, P(G) = .2 and P(P) = .15. I also know that P(G|P) = 4P(G| not P). I am looking for the conditional probability P(G| not P). I would imagine that I need to find P(G|P) / 4, but this would mean that I need $P(G \cap P) P(P)$ and alas, I do not believe that I have $P(G \cap P)$. ANy ideas? Anything I am missing? Thanks!

Note that

$$P(G) = P(G \cap P) + P(G \cap P^c) = P(G|P)P(P) + P(G|P^c) P(P^c)$$

You know everything except $P(G|P)$ and $P(G|P^c)$, and you know a relation between those two which should allow you to solve for $P(G|P)$.

I must be doing something wrong... I have

$.2 = (.15)P(G|P) + (.85)P(G|P^c) = (.15)P(G|P^c)/4 + (.85)P(G|P^c)$

and upon simplification I get $P(G|P^c) = .225$, but the answer is supposed to be .16.

Here is what I get. Write $x = P(G | P^c)$. Then

$$0.2 = (0.15) (4x) + (0.85) x = 1.45x$$

so

$$x = 0.2 / 1.45 = 0.1379...$$

P.S. You know .225 has to be wrong, since P(good grade) = 0.2 and surely the probability shouldn't go up for students who don't participate!

Thanks for all the help! It wouldn't be the first typo in the notes...