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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

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  • $\begingroup$ I think your Pr(Pick Right) is incorrect, the way you have it know you are assuming she doesn't know anything and is always guessing. The term you desire on the top is simply 0.7, more formally P(PR and Know) = P(PR | know) * Pr(know) = 1 * 0.7 = 0.7. $\endgroup$ – Grid Apr 4 '14 at 23:29
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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.

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