Combinations & Probability

In summary: You can use the union rule to add together the probabilities of these mutually exclusive outcomes to get the probability of "only one shift is represented".The same reasoning applies to part (d): The event "at least one shift is unrepresented" is the complement of "all three shifts are represented", which is the intersection of the three disjoint events "only the day shift is represented", "only the swing shift is represented", and "only the graveyard shift is represented".
  • #1
exitwound
292
1

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?
 
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  • #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".)
 
  • #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.
 
  • #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".
 
  • #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.
 
  • #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".
 

1. What is the difference between a combination and a permutation?

A combination is a selection of items where the order does not matter, while a permutation is a selection where the order does matter. For example, choosing three toppings on a pizza would be a combination, as the order in which the toppings are placed does not affect the final product. On the other hand, the order of a combination lock matters, making it a permutation.

2. How do you calculate the number of combinations?

The formula for calculating the number of combinations is nCr = n!/r!(n-r)!, where n is the total number of items and r is the number of items being chosen at a time. For example, if you have 5 different toppings to choose from and can only choose 3, the number of combinations would be 5C3 = 5!/3!(5-3)! = 10.

3. What is the "Probability Rule of Addition"?

The Probability Rule of Addition states that the probability of two or more events occurring is equal to the sum of their individual probabilities. This is represented by the equation P(A or B) = P(A) + P(B) - P(A and B), where P(A) and P(B) are the probabilities of the individual events and P(A and B) is the probability of both events occurring together.

4. How is probability related to combinations?

Probability is related to combinations through the calculation of the likelihood of an event occurring. Combinations are used to determine the total number of possible outcomes, which is then used to calculate the probability by dividing the number of desired outcomes by the total number of possible outcomes.

5. What is the "Probability Rule of Multiplication"?

The Probability Rule of Multiplication states that the probability of two or more independent events occurring together is equal to the product of their individual probabilities. This is represented by the equation P(A and B) = P(A) x P(B), where P(A) and P(B) are the probabilities of the individual events. This rule only applies to independent events, meaning that the outcome of one event does not affect the outcome of the other.

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