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Math Help - Probability.

  1. #1
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    Probability.

    Fifteen telephones have just been received at an authorized service center. Five are cellular, five are cordless, and five a conventional corded phones. If phones are randomly selected for service what is the probability that all the cordless phones are serviced first? What is the probability that after servicing ten phones, only two of the three types remain to be serviced?

    My Work

    Cellular - 5
    Cordless - 5
    Corded - 5

    Sorry I don't know how to put the problem in Math symbol.

    a) What is the probability that all the cordless phone are service first?
    (5C5)(10C1)/(15C3) = (1)(10)/455 = .02 or 2%

    b) What is the probability that after servicing ten telephones, only two of the three type remain to be service?
    All I can come up is (10C2)(___)/(15C3)
    Last edited by iMoons; February 1st 2010 at 02:21 PM.
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  2. #2
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    Hello, iMoons!

    Here's the first part . . .


    Fifteen telephones has just received at a service center.
    Five are cellular (A), five are cordless (B), and five conventional corded phones (C).

    (a) If phones are randomly selected for service, what is the probability
    . . .that all the cordless phones (B) are serviced first?
    There are: . {15\choose5,5,5} \,=\,756,\!756 possible orders for the 15 phones.


    There is one way for the B-phones to be first.

    The other 10 phones can follow in: . {10\choose5,5} \,=\,252 ways.

    Hence, there are: . 1\cdot252 \:=\:252 ways for the B-phones to be first.


    Therefore: . P(B\text{-phones first}) \;=\;\frac{252}{756,\!756} \;=\;\frac{1}{3003} \;\approx\;0.000333


    I'm still working on the second part . . .
    .
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  3. #3
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    I thought 15C3 = 455 or <br />
{15\choose 3} \,=455<br />
because 15 way out of 3 types.
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