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Ryan, Engineer

Category: Pre-Calculus

Satisfied Customers: 9046

Experience: B.S. in Civil Engineering

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I was trying to do the probability numbers on this equationThe

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I was trying to do the probability numbers on this equation The number of employees in the drug testing pool is 25k. So if 25% of the 25k employees, must be drug tested every year. Spread out over each quarter of the year. That is a total of 6,250 employees divided into 4 quarters. 1,562 employees would be selected each quarter. Then you must test 10% of the same 25k employees for alcohol. That is 2,500 more employees selected. Over 4 quarters that is 625 more employees selected. What is the probability of one employee being selected for both alcohol & drug test in the same quarter? Or how often could this type of event occur?

Submitted: 2 years ago.

Category: Pre-Calculus

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Customer:replied 2 years ago.

It might help if you could answer the question based on Leandro of time. For example if a employee worked under the above drug testing percentages. The avreage employee would be tested every 2yrs. So if you could say that the chances of bolth a drug & alcohol test would occur every 7yrs. Just a different way of thinking about it

Hi, Thank you for using the site. I'll be happy to help you with this. 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) = 0.001563 Converted 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. Please feel free to ask if you have any additional questions about this. Thanks, Ryan

Hi, The 0.16% value is the probability that any individual employee would be selected for both tests in a single quarter. You can also look at this as an individual employee being selected, on average, 16 times out of every 10,000 quarters. Ryan

Hi, Yes, 16 out of 10,000. Converting the percentage back to a decimal value gives: 0.16% = 0.0016 which is equal to 16/10,000. 16 times out of 1000 would be 0.016. Thanks, Ryan