document.write( "Question 589469: On a multiple-choice question with five choices, a certain student
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document.write( "either knows the answer and then marks the correct choice, or does not know the
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document.write( "answer and then marks one of the choices at random. What is the probability that he knew the answer if he marked the correct choice? Assume that the prior probability that he knew the answer is 3/4. \n" );
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Algebra.Com's Answer #374851 by richard1234(7193)![]() ![]() You can put this solution on YOUR website! Construct a table:\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "---------Know----Doesn't Know \n" ); document.write( "Correct 1-----------1/5 \n" ); document.write( "Incorrect 0-----------4/5\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Our sample space is limited to:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the student knew the answer and got it correct \n" ); document.write( "the student did not know the answer but got it correct.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The first case is easy, P(know) = 3/4 and P(correct) = 1, so P(know and correct) = 3/4.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "For the second case, P(doesn't know) = 1/4 and P(correct) = 1/5 (since there are 5 choices, one selected at random). Therefore P(doesn't know and correct) = (1/4)(1/5) = 1/20.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The probability that he got it correct is 3/4 + 1/20 = 13/20. However, we are only interested in P(know and correct) which is 3/4, or 12/20. Therefore, P(know | correct) = 12/13. \n" ); document.write( " |