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Math Help - Think this is an easy problem involving Bayes Theorem...

  1. #1
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    Think this is an easy problem involving Bayes Theorem...

    but I am not sure. It seems like the ball and urn type problem where Bayes Theorem can be easily applied but it is not quite the same. Can anybody help me out?

    A box contains 25 computer diskettes, 5 of which are double density and the remainder are high density. A second box contains 15 diskettes, 10 double density and 5 high density. A box is selected at random and then two diskettes are chosen randomly from the selected box. What is the probability that both diskettes are high density?
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  2. #2
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    Hello, babbagandu!

    Box 1 contains 25 computer diskettes: 5 DD and 20 HD.
    Box 2contains 15 diskettes: 10 DD and 5 HD.
    A box is selected at random and then two diskettes are chosen randomly from the selected box.
    What is the probability that both diskettes are HD?

    P(\text{box 1}) \:=\:\frac{1}{2}

    P(\text{both HD}) \;=\;\frac{20}{25}\cdot\frac{19}{24}\:=\:\frac{19}  {30}

    . . Hence: . P(\text{box 1}\wedge\text{both HD}) \:=\:\frac{1}{2}\cdot\frac{19}{30} \:=\:\frac{19}{60}


    P(\text{box 2}) \:=\:\frac{1}{2}

    P(\text{both HD}) \:=\:\frac{5}{15}\cdot\frac{4}{14} \:=\:\frac{2}{21}

    . . Hence: . P(\text{box 2} \wedge\text{both HD}) \:=\:\frac{1}{2}\cdot\frac{2}{21} \:=\:\frac{1}{21}


    Therefore: . P(\text{both HD}) \;=\;\frac{19}{60} + \frac{1}{21} \;=\;\boxed{\frac{51}{140}}

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  3. #3
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    Oh ok. That wasn't too bad, can't believe I didn't get that lol. Thank you for the help!
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