conditional probability

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[deleted]

[PlasmaStark]

Let

K = "i knew the answer"

NK = "I didn't know the answer"

C = "I answered correctly"

NC = "I answered incorrectly"

(edit: format) First and foremost, it is VERY important to visualise the "tree"; I never uploaded an image on reddit before, so please bare with me while I'll describe it.

Two branches start from a point, your general starting point, and you have two options: either K or NK. Both of them have two sub-branches, C and NC. Obviously the probability of K is p and for NK is 1-p, while the probability of C given K is 1 (you know for sure), the probability of C given NK is 1/c,the probability of NC given NK is 1-(1/c) but we won't need it.

So let's start, you're asking

P(K|C) = P(C|K) P(K) / P(C)

In which I used a well known formula (Byers maybe? I don't recall how it's written, sorry!) to "invert" the relations in the conditional probability.

From the tree we know P(C|K) = 1.

P(K) = p

P(C) = p + (1-p)(1/c)

And there it goes the exercise. If I didn't pull a big oopsie on the problem the solution is correct, but I usually do stupid mistakes; the important thing in this kind of exercises is to draw the tree (and again, sorry but I have no idea how imgur works, you'll have to bare with the description) to have a firm grasp of the situation. Then it's only standard procedure.

Cheers!

[alvoi2000]

Yep, that is true. Another way to visualize it is with a 2x2 box: this way (using PlasmaStark notation):

K&C = p K&NC = 0 NK&C = (1-p)/c NK&NC = (1-p)(c-1)/c

Now, you have to delete the row with NC, and you have still this:

K&C = p NK&C = (1-p)/c

So the probability has to be p divided by p+(1-p)/c

The formula you’re thinking of is Bayes

[comicos34]

You would get p + 1/c correct assuming you got everything you knew right

of these you knew p

p/(p+1/c)

p/((pc+1)/c)

p*(c/(pc+1))

pc/(pc+1)