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Homework Help: Permutations and Combinations of samples

  1. Jan 19, 2010 #1
    Review problem. I have the answers. Don't know how to get them. Answers in brackets.

    1. The problem statement, all variables and given/known data

    Shortly after being put into service, some buses manufactured by a certain company have developed cracks on the underside of the main frame. Suppose a paricular city has 25 of these buses, and cracks have actually appeared on 8 of them.

    a.)How many ways are there to select a sample of 5 buses from the 25 for a thorough inspection? [53130]

    b.) In how many ways can a sample of 5 buses contain exactly 4 with visible cracks? [1190]

    c.) If a sample of 5 buses is chosen at random, what is the probability that exactly 4 of the 5 will have visible cracks? [.0224]

    d.) If buses are selected as in part (c), what is the probability that at least 4 of those selected will have visible cracks? [.0235]

    2. Relevant equations

    [tex]Permutations = P_{k,n} = \frac{n!}{(n-k)!}[/tex]

    [tex]Combinations = \begin{pmatrix}
    n\\
    k
    \end{pmatrix}= \frac{n!}{k!(n-k)!}[/tex]

    3. The attempt at a solution

    a.) I'm looking for the total number of combinations of choosing 5 buses out of the 25 buses. Therefore

    [tex]Combinations = \begin{pmatrix}
    n\\
    k
    \end{pmatrix} = \frac{n!}{k!(n-k)!} = \frac{25!}{5!(25-5)!} = 53130[/tex]

    b.) I don't know how to set it up.
    c.) Didn't attempt it because I can't do (b).
    d.) Didn't attempt it because I can't do (b).
     
    Last edited: Jan 19, 2010
  2. jcsd
  3. Jan 19, 2010 #2

    vela

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    Your answer to part a is wrong. 25!/20! doesn't equal 53130; it's equal to 6375600.

    Do you understand the difference between permutations and combinations? With permutations, order matters; with combinations, it doesn't. For example, suppose you have n=3 items, A, B, and C and you select k=2 of them. There are 6=3!/(3-2)! permutations you can get, namely AB, BA, AC, CA, BC, and CB. But there are only 3=3!/[2!(3-2)!] combinations you can get because the pairs AB and BA, AC and CA, and BC and CB only count as one combination each since the order doesn't matter.

    In part a, the order of the busses doesn't matter, so you want to calculate the number of combinations, not permutations.
     
  4. Jan 19, 2010 #3
    Yes I figured that out but physicsforums went down and I couldn't edit it.
     
  5. Jan 19, 2010 #4

    vela

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    In part b, they're asking how many ways can you choose 5 buses from the 25 total such that 4 will contain cracks and one won't. That means you have to choose 4 from the 8 with cracks and 1 from the remaining 17.
     
  6. Jan 19, 2010 #5
    b.)[tex] \begin{pmatrix}
    8\\
    4
    \end{pmatrix} \begin{pmatrix}
    17\\
    1
    \end{pmatrix} = \frac{8!}{4!4!} * \frac{17!}{1!16!}= 1190[/tex]

    c.) [tex] \frac{\begin{pmatrix}
    8\\
    4
    \end{pmatrix} \begin{pmatrix}
    17\\
    1
    \end{pmatrix}}{\begin{pmatrix}
    25\\
    5
    \end{pmatrix}} = \frac{\frac{8!}{4!4!} * \frac{17!}{1!16!}}{\frac{25!}{5!20!}} = .0224[/tex]

    d.) [tex] \frac{\begin{pmatrix}
    8\\
    5
    \end{pmatrix} \begin{pmatrix}
    17\\
    0
    \end{pmatrix}}{\begin{pmatrix}
    25\\
    5
    \end{pmatrix}} + AnswerToPartC = \frac{\frac{8!}{5!3!} * \frac{17!}{0!17!}}{\frac{25!}{5!20!}} + .0224 = .0235[/tex]

    Is this how it's done?
     
  7. Jan 19, 2010 #6

    vela

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    Yup, perfect.
     
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