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Math Help - Probability Q12

  1. #1
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    Probability Q12 [ NEED HELP FOR PART C ]

    Attempt for a and b are fine. However c is wrong!!! require help for part C

    There are three cartons of vases, each containing 8 good vases and 2 defective ones. A buyer selects 2 vases at random from each carton to check. If both vases selected from a carton are good, he will accept the carton: otherwise, he will reject the carton. Find the probability that

    (a) he will accept a carton.
    (b) he will accept all three cartons.
    (c) he will accept at least one carton.

    Attempt

    a) P(he will accept a carton)= [\frac{8}{10}][\frac{7}{9}
    = \frac{28}{45}

    b)P(he will accept all three)= [\frac{28}{45}]^3
    = \frac{21952}{91125}

    c)P(at least 1 accepted)= 1-P(he will not accept any)
    = 1-(\frac{2}{10}\times\frac{1}{9})^3
    = 0.999
    Last edited by Punch; February 18th 2010 at 04:45 AM.
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  2. #2
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    P( he will not accept any ) = 1 - ( 2/90 )^3 = 0.999

    did you forgot to write that that or? ( than it is ok )
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  3. #3
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    Sorry, that was just a typo. In actual fact, I have already take the power of 3 into account.
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  4. #4
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    Does anybody have different thoughts/workings from me? Please post them!
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  5. #5
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    Quote Originally Posted by Punch View Post
    Does anybody have different thoughts/workings from me? Please post them!
    For (c), note that Pr(reject carton) = 1 - Pr(accept carton) = 1 - 56/90 = 34/90 ....
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