- #1
exitwound
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Homework Statement
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?
Homework 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]
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?