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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?

Probability
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use Bayes' formula?
\(P(\text{she knew|she got it right})=\frac{P(\text{got right|knew})}{P(\text{got right|knew})P(\text{knew})+P(\text{got right|didn't know})P(\text{didn't know})}\)
following.....

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Other answers:

for example, you know that the numerator is 1 for example you know that P(got it right|didn't know) is due to luck (like coin tossing..), it involves a 1/2 probability in here.
ok
now i think you can replace all quantities in the fraction. right?
yeah thank you
@reemii can you help me out in this question how to solve it
@reemii doesn't that give a probability greater than 1...?
i forgot the factor P(she knew) in the numerator. then it's correct.
Sounds like 14 times out of twenty she knows the answer 3 times out of twenty she guesses correctly 3 times out of twenty she does not. 14/17 she knew the answer.

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