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OpenStudy (anonymous):

On a true-false test, each question has exactly one correct answer: true, or false. A student knows the correct answer to 70% of the questions on the test. Each of the remaining answers she guesses at random, independently of all other answers. After the test has been graded, one of the questions is picked at random. Given that she got the answer right, what is the chance that she knew the answer? My first try was 0.70 * 1 + 0.3 * 1/2 = 85/100. But this is not correct. Could someone help me out ? Thank you!

OpenStudy (agent0smith):

In case the last link wasn't working correctly, http://openstudy.com/study#/updates/517e3410e4b007218446c5d5

OpenStudy (anonymous):

@agent0smith, I understand that the chance that she got the answer right is 0.85%, but given that she got the answer right, what is the chance that she knew the answer? I thought it's 0.7, but I was wrong. Could you help with this, please?

OpenStudy (agent0smith):

Its in the above link @njux

OpenStudy (agent0smith):

You still need to put the numbers i gave into the formula...

OpenStudy (anonymous):

Well, then I obviously don't get what in this case is A and what is B?

OpenStudy (agent0smith):

I didnt use A and B, i used K and R... i stated what they both are in the other question

OpenStudy (anonymous):

So this means that P (K|R) = P(R|K)*P(K)/P(R)?

OpenStudy (agent0smith):

Yep

OpenStudy (anonymous):

Thank you very much, everything is clear :)

OpenStudy (anonymous):

@agent0smith thanks a lot!

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