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Math Help - Conditional probability theorem (possiblying involving Bayes')

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    Conditional probability theorem (possiblying involving Bayes')

    A student either knows the answer or guesses on a multiple choice test.

    Let p be the probability the student knows the answer and 1-p be the probability the student guesses.

    Assume a student that guesses will be right with a probability of 1/m. (where m is the number of multiple choice alternatives.

    What is the probability that the student knew the answer to a question given that it was answered correctly?

    I am pretty sure this problem uses Bayes Thm, but I am not sure where to begin. Any help please? Thank you.

    -Jame
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    Quote Originally Posted by Jame View Post
    A student either knows the answer or guesses on a multiple choice test.

    Let p be the probability the student knows the answer and 1-p be the probability the student guesses.

    Assume a student that guesses will be right with a probability of 1/m. (where m is the number of multiple choice alternatives.

    What is the probability that the student knew the answer to a question given that it was answered correctly?

    I am pretty sure this problem uses Bayes Thm, but I am not sure where to begin. Any help please? Thank you.

    -Jame
    Draw a tree diagram and use it to insert the appropriate probabilities into the conditional probability formula. I get

    \displaystyle \frac{.....}{..... + \frac{1-p}{m}} = ....

    where the ..... are for you to fill in.
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    Thanks for answering.

    I partitioned the sample space into the events
    K - knowing the answering
    G - guessing the answering

    And I also have C, the event where the answer is correct
    C has probability 1 if you know the answer
    C has probability 1/m if you guess

    Therefore P(K|C) = \frac{P(K \cap C)}{P(C)} = \frac{1*p}{1*p + \frac{1-p}{m}}

    Is this correct?
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    Quote Originally Posted by Jame View Post
    Thanks for answering.

    I partitioned the sample space into the events
    K - knowing the answering
    G - guessing the answering

    And I also have C, the event where the answer is correct
    C has probability 1 if you know the answer
    C has probability 1/m if you guess

    Therefore P(K|C) = \frac{P(K \cap C)}{P(C)} = \frac{1*p}{1*p + \frac{1-p}{m}}

    Is this correct?
    Yes.
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