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I need to know the answer to a probability question.The

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I need to know the...
I need to know the answer to a probability question.
The total number of employees is 30,000
Annually 25% of the total employees will be selected randomly for a drug test.
Annually 10% of the total employees will be selected randomly for a alcohol test.
Annual drug & alcohol random selections will be spread out equally over the course of 4 quarters. Annual selections for drug and alcohol will be done independently of one another.
What is the probability of being selected for both drug and alcohol tests in the same quarter?
How many employees could be selected for both drug and alcohol on any given quarter in the course of a year?
looking over 8 years of data what would be the annual avreage of this type of selection?
Submitted: 2 years ago.Category: Calculus and Above
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6/17/2015
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
Ray Atkinson
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The fact that it is stated that the selections are independent is important and makes the problem much easier. The probability of two independent choices happening is the product of the individual probabilties, so here it is 1/4 * 1/10 = 1/40
Multiply that by 30000 and the expected value is that 750 will get picked for both tests in a given year.
Since each year is also independent, the average is going to be about 750 each of the 8 years.
(I just ran a 100-year simulation of this and got(###) ###-####per year. "That's close enough for government work," as my grandfather used to say.)
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Customer reply replied 2 years ago
This does not seem right. You have not calculated the fact that the selections are spread out over the course of the year. So each quarter there is a selection of 1/4 the annual amount. So I need to know what the chances are of being selected for both. Then having both selections fall in the same quarter of the year.
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
Oh, you're right. Let me figure that for you.
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Customer reply replied 2 years ago
10% selected for alcohol annually spread out over 4 quarters. Then 25% selected for drugs annually spread out over 4 quarters.So a employee would have to get selected in the 10% alcohol & then selected in the 25% drug. Then get selected in the same quarter.
Customer reply replied 2 years ago
When I say independently of one another I mean that each selection is drawn from the annual total # ***** 30k
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
Let's look at it this way. They make a list at the beginning of the year with everyone to be tested on it. There is a 1/4 chance that you are on the drug list, and 1/4 of those are on it for the first quarter, so a 1/16 chance. There is also a 1/10 chance that you are on the alcohol list, and a 1/4 chance that it is for the first charter, so a 1/40 chance. This means (again, independent) a 1/640 chance, or 46.875 people out of 30000 will be on both lists in a quarter, so that is 187.5 per year.
Running it through my simulation, I get 46.81188 and 46.55445, so the empirical numbers match, too.
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Customer reply replied 2 years ago
Still not right.30k em
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
The numbers work out. Could the issue be rounding off?
1/4 * 1/4 * 1/10 * 1/4 = 1/640 (0.0015625 = 0.15625%)
1/640 * 30000 = 46.875 (could it be 47 rounded?)
Average per year is 187.5, or 188?
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Customer reply replied 2 years ago
This is the answer I got from another member using 25k as the total number of employees tested.If 25% of the employees are selected annually for drug testing, and are evenly distributed over all four quarters of the year, then the probability of being selected for a drug test in a quarter would be 0.25 / 4 = 0.0625.This is because any one employee has a 25% chance of being selected during the whole year, and 1/4 of that percentage each quarter.You will get approximately the same percentage by dividing the 1562 employees that you calculated by the 25,000 total employees. There will be some rounding differences since the 6250 annual selections cannot be evenly divided into four equal amounts for each quarter.The alcohol testing is calculated in a similar manner:10% per year / 4 quarters = 2.5% per quarter, which is equal to a probability of 0.025.Now, assuming that the events are independent (meaning that whether an employee was selected for drug testing has no effect on them being selected for alcohol testing), then the probability of BOTH happening is equal to the product of the individual probabilities:P(both drug AND alcohol testing) = P(drug test) * P(alcohol test)= (0.0625)(0.025) = 0.001563Converted to a percentage then, there is about 0.16% chance of being selected for BOTH tests in any one quarter.If the results of one test had some effect on whether an employee was selected for the other test (for instance, if failing a drug test required that the employee also be tested for alcohol), then the above probability would not be accurate. In this case, the two events aren't independent.
Customer reply replied 2 years ago
So how many employees would be selected for both test in the same quarter in one year?
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
That other person gave the same numbers. (0.0625)*(0.025) = 0.0015625
0.0015625 * 30000 = 46.875 per quarter, or 187.5 per year. The problem must be rounding error.
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Customer reply replied 2 years ago
Why would you take those numbers and multiply again by 30kYou just took the 0.0015625 and multiplied it by 30k. Why would you do that when 30k is the total number of employees tested and 0.0015625 is the probability.
Customer reply replied 2 years ago
Let me see if I make sense of this.
30,000 employes in anual selection
25% selected for drug test.
That is 7,500 employees selected.
Equally divided into 4 quarters.
That is 1875 employees per quarter selected for a drug test.Then you have 30k employees selected for alcohol at 10%
That equals 750 selections per quarter.Now I need some help. If there is a 25% chance of drug selection and a 10% chance of alcohol selection. Then what is the probability of being selected for both? Then what is the probability of being selected for both in the SAME quarter?Let's say 1875 employees get selected drug test in Q1. Then 750 employees get selected alcohol test in Q1. Then you have to figure the ods of one employee being selected for BOTH test in Q1. NOT the probability of being selected for both test annually.
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
Ummmm, yeah. You could do the numbers that way, but why?
What are the chances that you are ones of the 1875? 1875/30000=0.0625
What are the chances that you are ones of the 750? 750/30000=0.025
What is the probability of being selected for both drug and alcohol tests in the same quarter? (0.0625)(0.025) = 0.0015625
How many employees could be selected for both drug and alcohol on any given quarter in the course of a year? 0.0015625 * 30000 = 187.5 ≈ 188
Looking over 8 years of data what would be the annual avreage of this type of selection? It would be the same as one year's average.
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Customer reply replied 2 years ago
I need a new expert. You are completely wrong. Once again you took the resukts and multiplied by 30k. That is just the lain wrong. I'm sorry I need someone else to answer this.
Tutor: Ray Atkinson, Bachelor's Degree replied 2 years ago
Of course I took it times the 30k. If something happens to 1/640 of a population and you want to know how many people if affects, you multiply the 1/640 times the population.
Best of luck.
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Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
Mr. Glenn
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Customer reply replied 2 years ago
Is this good my to be another charge on my accounts by? I see that I have billed already for what I belive is a incite t answer to my question.
Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
Hi!
As long as you don't yet rate positively (ok, good, or excellent service) the answers posted, then you are not yet billed or your deposit is not yet deducted.
By the way,
What time or how many hours/minutes do you need the answers?
I am now working on these.
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Customer reply replied 2 years ago
In the next several hours. I need accuracy. I have some conflicting answers.
Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
Ok
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Customer reply replied 2 years ago
Any results?
Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
not yet sir,
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Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
Hi there!
Here are the correct answers:
1.)
What is the probability of being selected for both drug and alcohol tests in the same quarter?
solution:
= 0.25(0.1)*(4/16) = 1 / 160 or 0.00625 or 0.625%
2.) How many employees could be selected for both drug and alcohol on any given quarter in the course of a year?
Since this is a simple binomial distribution, the average number of employees could be selected for both drug and alcohol on any given quarter in the course of a year = np = 30000(1/160) = 187.5 employees (or approximately 88 employees)
3.)
looking over 8 years of data what would be the annual average of this type of selection
solution:
= (187.5 + 187.5 + 187.5 + 187.5 + 187.5 + 187.5 + 187.5 + 187.5 )/8
= 187.5 employees
(annual average for 8 years will be just the same as the answer for Q2)
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Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
Hi
There is a slight typo with the answer for #2
here it is:
187.5 employees (or approximately 188 employees)
(88 should be 188)
Are you satisfied now with the answers? Do you still have some more math questions?
Thanks
Regards
Mr. Glenn
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Customer reply replied 2 years ago
Just to clarify. I see you did the calculations in the first line.
= 0.25(0.1)*(4/16) = 1 / 160 or 0.00625 or 0.625%
Did you calculate for 10% alcohol?
I see you used 0.25 I would think that is for the 25% chance of a drug test. I don't see where you combined the odds of both the 10% & 25% then 1/4
Tutor: Mr. Glenn, Math and Stat Expert replied 2 years ago
The answer to your clarification why I only used 4/16 and not 1/16 is that the question is
"What is the probability of being selected for both drug and alcohol tests in the same quarter?"
Since it is not specified which among the 4 quarters does the two testing will be done, then there are 4 possible quarter it can happen, that is why we need to multiply it by 4, but if the question is
"What is the probability of being selected for both drug and alcohol tests in the same first quarter ?"
then instead of using 4/16, I will now just use 1/16 since it is now specified which among the 4 quarters the two testings will happen.
So that is the reason 4/16 is more correct factor than 1/16 since specificity of the quarter is not stated.
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