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Math Help - another urn probability

  1. #1
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    another urn probability

    An urn contains 2 fair coins and one 2-headed coin. A coin is drawn randomly and tossed twice. Given that both tosses resulted in H, find the conditional probability that the 2-headed coin was chosen (as a fraction in lowest terms). HINT: use Baye's Formula

    Next: Repeat problem, assuming the urn contains 8 fair coins and one 2-headed coin.
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  2. #2
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    Quote Originally Posted by mhitch03 View Post
    An urn contains 2 fair coins and one 2-headed coin. A coin is drawn randomly and tossed twice. Given that both tosses resulted in H, find the conditional probability that the 2-headed coin was chosen (as a fraction in lowest terms). HINT: use Baye's Formula

    Next: Repeat problem, assuming the urn contains 8 fair coins and one 2-headed coin.
    Define the event A=Draw the 2-headed coin. B=get 2 heads in 2 tosses.

    Bayes formula tells us that

    P(A|B)=\frac{P(B|A)P(A)}{P(B|A)P(A)+P(B|A^c)P(A^c)  }

    Now we can figure out all of the probabilties on the right hand side

    P(B|A)P(A)=1\cdot \frac{1}{3}

    P(B|A^c)P(A^c)=\frac{1}{4}\cdot \frac{2}{3}

    From here just plug into the formula
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