There are 4 alternatives on a multiple choice test. Let suppose that a student learned 70% of the material. If she doesn't know the answer, she picks one randomly.
If she picked the right answer, what is the probability that she actually knew the answer.
What I did so far:
$P(Knows)=0.7$
$P($PicksRight)=0.25
$P(Knows|PicksRight)=\frac{P(PR\cap K)}{P(PR)}$
I don't know how I get the $P(PR\cap K)$ probability.
Thanks in advance
Take a particular question. Let $K$ be the event she knows the right answer. Let $C$ be the event she answers the question correctly.
We want $\Pr(K|C)$, which is $\frac{\Pr(K\cap C)}{\Pr(C)}$.
She can be right in two ways: (i) She knows the right answer, and of course picks it or (ii) She doesn't know, but guesses correctly.
The probability of (i) is $0.7$, or if you prefer, $(0.7)(1)$.
The probability of (ii) is $(0.3)(0.25)$.
For $\Pr(C)$, add the probabilities for (i) and (ii).
The probability of $K\cap C$ is easier, it is $0.7$.
Now you have all the ingredients.