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Math Help - Pthat a bunny came out of Magician's green hat!?!

  1. #1
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    Pthat a bunny came out of Magician's green hat!?!

    So here goes the question:

    A magician has two hats containing flowers and animals.
    The black hat has 3 roses and 2 bunnies.
    The green hat has 5 roses and 3 bunnies.

    A volunteer from the audience rolls two standard dice (numbers 1 to 6). If the sum of the numbers is equal to or less than 4, the magician takes an object out from the green hat; otherwise the magician takes an object from the black hat.

    If the magician obtains a bunny,
    what is the probability that it came from the green hat?
    I've approached this question with the presumption that the two events are independent. For the magician to have even considered putting his hand in the green hat; there had to be a sum of 4 or less produced by the rolling of the two dice.
    I calculated the chance of rolling sum of 4 or less to be: 6/36 = 0.1666

    The chance that a bunny is pulled out can be calculated by: 3/8 = 0.375


    adding the above two values reveals the probability to be : 0.5416


    What do you guys think? If my reasoning flawed? Are these mutually dependent events? Any insight would be greatly appreciated!
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  2. #2
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    If my reasoning flawed?
    Yes. The question is, GIVEN that a bunny has been drawn, what is the probability that it came from the green hat. This is an application of Bayes Theorem.

    Consider the entire sample space to be of items drawn. He'll either draw from Hat Black or Hat Green. The probability that he draws from Hat Black is 30/36 (or the compliment of the probability that he draws from hat green, which you already calculated). The probability that he draws from Hat Green is 6/36. Now, OF the items in Hat Black 2/5 of them are bunnies. OF the items in Hat Green, 3/8 of them are bunnies. Therefore the probability that he draws a bunny GIVEN that he is drawing from Hat Black is (2/5)(30/36), and the probability that he draws a bunny GIVEN that he is drawing from Hat Green is (3/8)(6/36).

    What YOU have to figure out, is GIVEN that he has drawn a bunny (we are reversing our "given" statements), what is the probability that he drew it from Hat Green? So what you have to do is consider the new sample space. In this case the "sample space" will be "drawing a bunny" (remember, that the GIVEN part of these problems defines the new sample space you are working from).
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  3. #3
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    Red face

    Quote Originally Posted by ANDS! View Post
    Yes. The question is, GIVEN that a bunny has been drawn, what is the probability that it came from the green hat. This is an application of Bayes Theorem.

    Consider the entire sample space to be of items drawn. He'll either draw from Hat Black or Hat Green. The probability that he draws from Hat Black is 30/36 (or the compliment of the probability that he draws from hat green, which you already calculated). The probability that he draws from Hat Green is 6/36. Now, OF the items in Hat Black 2/5 of them are bunnies. OF the items in Hat Green, 3/8 of them are bunnies. Therefore the probability that he draws a bunny GIVEN that he is drawing from Hat Black is (2/5)(30/36), and the probability that he draws a bunny GIVEN that he is drawing from Hat Green is (3/8)(6/36).

    What YOU have to figure out, is GIVEN that he has drawn a bunny (we are reversing our "given" statements), what is the probability that he drew it from Hat Green? So what you have to do is consider the new sample space. In this case the "sample space" will be "drawing a bunny" (remember, that the GIVEN part of these problems defines the new sample space you are working from).

    Thanks for the explanation ANDS!

    so does that mean the answer is just [6/36] = 0.1667?
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  4. #4
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    Remember now, the given part of our statement will determine our sample space. We are being told that the bunny has already been drawn. The TOTAL probability of drawing a bunny is the probability of drawing a bunny from the green hat, and the probability of drawing a bunny from the black hat. These are two mutually exclusive exhaustive events correct (since it is impossible for someone to throw a dice and be able to draw both from the green AND black hat). So we have to figure out the total probability of drawing a bunny. That is what we did:

    (2/5)(30/36)+(3/8)(6/36).

    Using Bayes Theorem:

    P(GH|B)=\frac{P(B|GH)P(GH)}{P(B|GH)P(GH)+P(B|BH)P(  BH)}

    Where, P(GH|B) is the probability of drawing from the green hat, given you drew a bunny, P(B|GH) is the probability of drawing a bunny given you drew from the green hat, and P(B|BH) the probability of drawing a bunny given you drew from the black hat.

    An answer of 6/36 is simply P(B|GH), which is not what we are looking for. Use the above equation to see if you can finish this one up.
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