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Math Help - combinations

  1. #1
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    combinations

    Suppose that three computer boards in a production run of forty are defective. A sample of five is to be selected to be checked for defects.
    a. How many different samples can be chosen?
    b. How many samples will contain at least one defective board?
    c. What is the probability that a randomly chosen sample of five contains at least one defective board?

    A) \binom{40}{5} = \frac{40!}{5!(35!)} = 658\;008 it didn't state something along the lines "{1,2,3,4,5} is different from {1,4,3,2,5}" so I am assuming its not a permutation question

    B) \binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25

    C) P(E) = \frac{N(E)}{N(S)} = \frac{25}{658\;008}
    Last edited by Jonroberts74; July 29th 2014 at 04:27 PM.
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  2. #2
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    Re: combinations

    Hello, Jonroberts74!

    Suppose that three computer boards in a production run of forty are defective.
    A sample of five is to be selected to be checked for defects.
    a. How many different samples can be chosen?
    b. How many samples will contain at least one defective board?
    c. What is the probability that a randomly chosen sample of five contains at least one defective board?

    (a)\;\binom{40}{5} = \frac{40!}{5!(35!)} = 658\,008 . Correct!

    (b)\;\binom{5}{1}+\binom{5}{2}+\binom{5}{3} = 25 . This is incorrect.

    We have: . \begin{Bmatrix}\text{3 def.} & (D) \\ \text{37 good} & (G)\end{Bmatrix}

    This is what you want: . \displaystyle\begin{Bmatrix} \text{1D,4G} & \text{2D,3G} & \text{3D,2G} \\ {3\choose1}{37\choose4} & {3\choose2}{37\choose3} & {3\choose3}{37\choose2} \end{Bmatrix}
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  3. #3
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    Re: combinations

    Quote Originally Posted by Soroban View Post
    Hello, Jonroberts74!


    We have: . \begin{Bmatrix}\text{3 def.} & (D) \\ \text{37 good} & (G)\end{Bmatrix}

    This is what you want: . \displaystyle\begin{Bmatrix} \text{1D,4G} & \text{2D,3G} & \text{3D,2G} \\ {3\choose1}{37\choose4} & {3\choose2}{37\choose3} & {3\choose3}{37\choose2} \end{Bmatrix}
    \binom{3}{1}\binom{37}{4}+\binom{3}{2}\binom{37}{3  }+\binom{3}{3}\binom{37}{2} = 222\,111

    C) P(E) = \frac{N(E)}{N(S)} = \frac{222\,111}{658\,008} \approx 33.76\%
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