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Math Help - multiple choice questions

  1. #1
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    Question multiple choice questions

    hello can anyone help answer this tricky question?

    multiple choice questions

    If you have a test with 6 questions and 4 possible answers to each
    question , what is probability i get 2 correct out of the six questions?


    i think the answer is 1215 - 4096
    but im not sure can any 1 verify

    its doing my head in!!!!!!

    tx
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  2. #2
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    Quote Originally Posted by jickjoker View Post
    hello can anyone help answer this tricky question?
    multiple choice questions
    If you have a test with 6 questions and 4 possible answers to each
    question , what is probability i get 2 correct out of the six questions?
    I am not quite sure what this is asking.
    First. Are you just guessing at the answer on each question?

    Second. Do you mean gatting exactly two questions correct or do you mean at least two questions? Which is it?

    Please try to clairfy this question.
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  3. #3
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    Are you saying that there is a 1215/4096 chance of you getting two answers correct out of six. Also I am assuming you are GUESSING to what the answers are.

    In the case that you are guessing, then this is a binomial distribution problem:

    The question involved "n" trials (the trials being individual questions).
    Each of the trials involves only two outcomes; the question is either RIGHT or it is WRONG. The individual probabilities for each equation is p=.25 and q=.75.
    Getting a right answer on one question is not going to affect getting a right answer on another.

    So essentially you need to know the total number of possible combinations of "six choose 2" that you can make from this. The equation would then be:

    [6!/2!(4!)][p^(.25)q^(.75)] where 6 is the number of trials (questions answered), 2 is the number you want to get right, 4 is (6-2, or n-k, where n is the number of trials and k is the number you of "successes"), p is the probability of getting a success in an individual trial and q is the probability of failure.
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  4. #4
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    im saying you get exactly 2 questions right from guessing
    what is the probability you get exactly 2 right
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  5. #5
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    Quote Originally Posted by jickjoker View Post
    im saying you get exactly 2 questions right from guessing
    what is the probability you get exactly 2 right
    \binom{6}{2}\left(\frac{1}{4}\right)^2\left(\frac{  3}{4}\right)^4 where \binom{6}{2}=\frac{6!}{(2!)(4!)}
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  6. #6
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    Quote Originally Posted by Plato View Post
    \binom{6}{2}\left(\frac{1}{4}\right)^2\left(\frac{  3}{4}\right)^4 where \binom{6}{2}=\frac{6!}{(2!)(4!)}
    I had to copy and paste someone's answer to type this way - where would one go to get a list of commands to type in this manner?
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  7. #7
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    Quote Originally Posted by ANDS! View Post
    I had to copy and paste someone's answer to type this way - where would one go to get a list of commands to type in this manner?
    Go to this forum
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  8. #8
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    im not sure about your answers

    somone else asked a similar question

    If you have a test with 6 questions and 4 possible answers to each
    question, and you guess all answers, what is the probability of
    getting all 6 questions right?

    1/4 x 1/4 x 1/4 x 1/4 x 1/4 x 1/4 = (1/4)^6 = 1/4096


    A more difficult problem would be to figure out the probability of
    getting 3 out of the 6 correct.

    Suppose you got the first three correct and the last three wrong.
    We already know the probability of getting one question right: 1/4.
    This means the probability of getting one question wrong must be 3/4
    (you have to either get it right or wrong - there's no middle ground
    here). The probability of this particular sequence of right and wrong
    answers would be:
    1/4 x 1/4 x 1/4 x 3/4 x 3/4 x 3/4 = (1/4)^3 (3/4)^3

    However there are many possible sequences for right and wrong answers
    giving 3 correct and 3 incorrect. This is where the terms of Pascal's
    triangle would come in. The number of possible sequences in this case
    is given by:
    6.5.4
    6_C_3 = ------- = 20
    1.2.3
    So there are 20 possible sequences giving three correct and three
    incorrect answers.
    The probability of exactly three correct answers is:
    20(1/4)^3 (3/4)^3
    = 540/4096
    = 135/1024


    ive tried to work out the answer the same way this is has been
    and i get 1215 /4096

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  9. #9
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    That is exactly what this:

    \binom{6}{2}\left(\frac{1}{4}\right)^2\left(\frac{  3}{4}\right)^4

    is. The first brackets are the total number of ways you can arrange 2 right answers amongst 6 total questions. The second brackets is your success probability raised to the number of correct answers (or (1/4*1/4) and the final brackets is the probability of getting a wrong answer raised to the number of wrong answers (or 3/4*/3/4*3/4*3/4). All that equation is, is a tree-diagram reduced to a simple formula.
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