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Thread: Combination Problems

  1. October 15th 2007, 07:47 AM #1
    Nellie
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    Combination Problems

    there are 12 resistors. 3 are defective. if you choose 4 what is the probability that you'll get:

    no defective resistors
    1 defective resistor
    3 defective resistors

    15 insurance policies: 8 life, 5 auto, 2 home. if you choose 3, what is the probability that you'll get
    all life policies
    both home policies
    1 of each
    all auto policies
    2 life and 1 auto policy

    Help please.
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  2. October 15th 2007, 08:43 AM #2
    Soroban
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    Hello, Nellie!

    There are 12 resistors; 3 are defective.
    If you choose 4 what is the probability that you'll get:

    (a) no defective resistors . (b) 1 defective resistor . (3) 3 defective resistors
    There are: . {12\choose4} = 495 possible samples.


    (a) "no defective" means all four are good.
    There are: . {9\choose4} = 126 ways to choose four good resistors.
    Therefore: . P(\text{0 d{e}f}) \:=\:\frac{126}{495} \;=\;\frac{14}{55}


    (b) "one defective" means 1 defective and 3 good resistors.
    There are: . {3\choose1}{9\choose3} \:=\;3\cdot84 \:=\:252 ways.
    Therefore: . P(\text{1 d{e}f}) \:=\:\frac{252}{495} \;=\;\frac{28}{55}


    (c) "3 defective" means all 3 defective resistors and 1 good resistor.
    There are: . {3\choose3}{9\choose1} \:=\:1\cdot9 \:=\:9 ways.
    Therefore: . P(\text{3 d{e}f}) \;=\;\frac{9}<br /> {495} \;=\;\frac{1}{55}
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  3. October 15th 2007, 09:15 AM #3
    Nellie
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    Soroban

    Thank you! and you explained it so clearly!
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