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Homework Help: Combinations & Probability

  1. Jan 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A production facility employs 20 workers on the day shift, 15 workers on the swing shift, and 10 workers on the graveyard shift. A quality control consultant is to select 6 of these workers for in-depth interviews. Suppose the selection is made in such a way that any particular group of 6 workers has the same chance of being selected as does any other group (drawing 6 slips without replacement from among 45).

    a.) How many selections result in all 6 workers coming from the day shift? What is the probability that all 6 selected workers will be from the day shift?

    b.) What is the probability that all 6 selected workers will be from the same shift?

    c.) What is the probability that at least two different shifts will be represented among the selected workers?

    d.) What is the probability that at least one of the shifts will be unrepresented in the sample of workers?

    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.) Are they asking for permutations? It doesn't really say whether or not in the question. I assume that it's asking for every way that 6 members can be chosen from the day shift. If that's the case then am I right with:

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

    I'm first finding the number of ways to get 6 out of the 20 day workers, then multiplying that by taking 0 more out of the remaining 25, correct?

    [tex]
    Outcomes = ^6\mathbb{P}_{20} * ^0\mathbb{P}_{25} = 27,907,200
    [/tex]

    The probability of all 6 being from the day shift would be:
    {Permutations of 6 from the 20 day shifts} / {Permutations of taking 6 from the total 45}

    [tex]\frac{^6\mathbb{P}_{20} * ^0\mathbb{P}_{25}}{^6\mathbb{P}_{45}} = 4.76x10^{-3}[/tex]

    Am I close on this one?

    b.) In order to figure this out, we need to find the probability that 6 come from the day shift, the probability that 6 come from the swing shift, and the probability that 6 come from the graveyard shift.

    Prob = P(Day) + P(Swing) + P(Graveyard)

    P(Day) = answer to a.)
    P(Swing) = [tex]\frac{^6\mathbb{P}_{15} * ^0\mathbb{P}_{25}}{^6\mathbb{P}_{45}} = 6.15x10^{-4}[/tex]
    P(Graveyard) = [tex]\frac{^6\mathbb{P}_{10} * ^0\mathbb{P}_{25}}{^6\mathbb{P}_{45}} =2.58x10^{-5}[/tex]

    How is this so far?
     
  2. jcsd
  3. Jan 21, 2010 #2
    The wording of the question does not make clear that order is significant, so you should assume it is not (i.e., use combinations rather than permutations). Your strategy is otherwise correct, and the probability numbers (not the numbers of distinct outcomes) are actually left unchanged by the choice of permutations versus combinations, as you can verify by expanding into factorials.

    A few TeX tips: Use \times for a cross-shaped multiplication sign [tex]a \times b[/tex], \cdot for a multiplication dot [tex]a \cdot b[/tex]; don't use an asterisk. If you really want the superscript number before the P, you want to stick it to the P instead of the multiplication sign: write {}^6 P_{10} \cdot {}^0 P_{25} to make [tex]{}^6 P_{10} \cdot {}^0 P_{25}[/tex]. Note the spacing. ({} is "nothing".)
     
  4. Jan 21, 2010 #3
    For part (c)...

    Would I be right in using:

    probability that at least two different shifts will be represented = (1 - Probability of 1 from each shift)?

    And in finding that, I'd use

    [itex]1 - ({}^1P_{20} + {}^1P_{15} + {}^1P_{10})[/itex]



    For part (d), I'm having trouble coming up with the way to attack this.
     
  5. Jan 21, 2010 #4
    No. The opposite of "at least two different shifts are represented" is "only one shift is represented". The opposite of "at least one shift is unrepresented" is "all three shifts are represented".
     
  6. Jan 21, 2010 #5
    But then I can't answer (c) with a single number. The probability of a day shift being represented is not the same as the probability of a swing shift, or a graveyard shift. The three probabilities are different aren't they?? I have a *really* hard time converting the wording of these problems into mathematical ideas.
     
  7. Jan 22, 2010 #6
    The trick here is that "only one shift is represented" is the union of three distinct, disjoint events -- which, as you point out, do have different probabilities. The event "only one shift is represented" is the union of "only the day shift is represented", "only the swing shift is represented", and "only the graveyard shift is represented".
     
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