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pakinam

  • one year ago

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?

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  1. reemii
    • one year ago
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    use Bayes' formula?

  2. reemii
    • one year ago
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    \(P(\text{she knew|she got it right})=\frac{P(\text{got right|knew})}{P(\text{got right|knew})P(\text{knew})+P(\text{got right|didn't know})P(\text{didn't know})}\)

  3. pakinam
    • one year ago
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    following.....

  4. reemii
    • one year ago
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    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.

  5. pakinam
    • one year ago
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    ok

  6. reemii
    • one year ago
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    now i think you can replace all quantities in the fraction. right?

  7. pakinam
    • one year ago
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    yeah thank you

  8. sarah10
    • one year ago
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    @reemii can you help me out in this question how to solve it

  9. agent0smith
    • one year ago
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    @reemii doesn't that give a probability greater than 1...?

  10. reemii
    • one year ago
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    i forgot the factor P(she knew) in the numerator. then it's correct.

  11. annacn1024
    • one year ago
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    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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