/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Many companies are now testing p... [FREE SOLUTION] | 91Ó°ÊÓ

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Many companies are now testing prospective employees for drug use. However, opponents claim that this procedure is unfair because the tests themselves are not \(100 \%\) reliable. Suppose a company uses a test that is \(98 \%\) accurate- that is, it correctly identifies a person as a drug user or nonuser with probability \(.98-\) and to reduce the chance of error, each job applicant is required to take two tests. If the outcomes of the two tests on the same person are independent events, what are the probabilities of these events? a. A nonuser fails both tests. b. A drug user is detected (i.e., he or she fails at least one test). c. A drug user passes both tests.

Short Answer

Expert verified
Short Answer: The probabilities are as follows: a) The probability that a nonuser fails both tests is 0.0004. b) The probability that a drug user is detected (i.e., they fail at least one test) is 0.0396. c) The probability that a drug user passes both tests is 0.9604.

Step by step solution

01

Identify the probabilities of failure for a nonuser:

Since the test is \(98 \%\) accurate, the probability that a nonuser passes the test is \(0.98\), and therefore the probability that a nonuser fails the test is \(0.02\) (because \(1 - 0.98 = 0.02\)).
02

Calculate the probability of a nonuser failing both tests:

Since the outcomes of the tests are independent events, we can use the formula we mentioned earlier to find the probability of a nonuser failing both tests: \(P(\text{nonuser fails both tests}) = P(\text{nonuser fails first test}) * P(\text{nonuser fails second test}) = 0.02 * 0.02 = 0.0004\) So, the probability that a nonuser fails both tests is \(0.0004\). #b. A drug user is detected (i.e., they fail at least one test).#
03

Identify the probabilities of failure and passing for a drug user:

Similar to the nonuser case, the probability that a drug user passes the test is \(0.98\), while the probability they fail the test is \(0.02\).
04

Calculate the probability of a drug user passing both tests:

Again, since the outcomes of the tests are independent events, we can find the probability of a drug user passing both tests using the formula: \(P(\text{drug user passes both tests}) = P(\text{drug user passes first test}) * P(\text{drug user passes second test}) = 0.98 * 0.98 = 0.9604\)
05

Calculate the probability of a drug user failing at least one test:

We can find the probability of a drug user failing at least one test by subtracting the probability of them passing both tests from \(1\): \(P(\text{drug user fails at least one test}) = 1 - P(\text{drug user passes both tests}) = 1 - 0.9604 = 0.0396\) So, the probability that a drug user is detected (i.e., they fail at least one test) is \(0.0396\). #c. A drug user passes both tests.# We already calculated the probability that a drug user passes both tests in step 2 of part (b): \(0.9604\). Therefore, the probability that a drug user passes both tests is \(0.9604\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Events
Understanding independent events is crucial in probability theory, especially when determining the likelihood of multiple occurrences. Two events are considered independent if the outcome of one does not affect the outcome of the other. For example, when a company administers drug tests for potential employees, the results of these tests may be treated as independent events if the result of one test does not influence the result of another.
In the given problem, each job applicant takes two separate drug tests. The tests are defined as independent events because the outcome of one test (pass or fail) does not alter the outcome of the other.
  • If the probability of passing one test is 0.98, then the chance of passing both tests is the product of each test’s probability: \( P(\text{pass both}) = 0.98 \times 0.98 = 0.9604 \).
  • Similarly, the probability of failing both tests is determined by multiplying the probability of failing one test with itself.
This understanding of independence allows us to calculate the probability of compounded events by simply multiplying their probabilities.
Testing Accuracy
Testing accuracy plays a critical role in determining the reliability of test results. Accuracy in this context refers to the probability that a test correctly identifies a drug user or nonuser.In the problem, we have a testing accuracy of 98%. This means that there is a 98% chance that the test will identify a nonuser correctly as not using drugs, and a 98% chance that it will identify a drug user correctly.
However, there is still a 2% chance of making an incorrect identification—where a nonuser fails the test (false positive) or a drug user passes the test (false negative). These inaccuracies, although small, are significant in scenarios involving human decisions.
  • A test with an accuracy of 98% means for a nonuser: \( P(\text{nonuser fails}) = 1 - 0.98 = 0.02 \).
  • For a drug user, the same 98% applies inversely: \( P(\text{drug user passes}) = 0.02 \).
This measure of testing accuracy helps in calculating realistic scenarios, such as the chance of a nonuser failing both tests or determining if a drug user will pass or fail at least one test.
Complement Rule
The complement rule is an incredibly helpful probability principle that allows us to determine the probability of an event not happening. According to this rule, the probability of an event occurring, plus the probability of it not occurring, equals one.Using the complement rule can simplify calculations, especially in probability exercises like drug testing.
For instance, suppose we want to find the probability that a drug user fails at least one test. After finding the probability of a drug user passing both tests, which is \(0.9604\), we can use the complement rule to find the chance of failing at least one test.
  • According to the complement rule, \( P(\text{fail at least one}) = 1 - P(\text{pass both}) \).
  • Thus, \( P(\text{fail at least one}) = 1 - 0.9604 = 0.0396 \).
By using the complement rule, calculations become more efficient by focusing on what we know and using that information to deduce what we need.

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