Composite Drug Test Based on the data in Table 4-1, assume tha...

Question

Composite Drug Test Based on the data in Table 4-1, assume that the probability of a randomly selected person testing positive for drug use is 0.126. If drug screening samples are collected from 5 random subjects and combined, find the probability that the combined sample will reveal a positive result. Is that probability low enough so that further testing of the individual samples is rarely necessary?

Accepted Answer

Welcome back, everyone. Suppose that the probability that a randomly selected employee's security badge scan fails at the entrance is 0.074. A company's system checks badge scans in groups of 5 employees. The system flags the group only if at least one scan fails. Find the probability that the group of 5 employees will trigger a flag due to at least 1 batch failure should the company prepare for frequent individual rechecks. So first of all, we're going to introduce a random variable X which represents the number of failures. What we want to do is basically identify the probability of X being greater than or equal to 1, meaning at least one failure. And we're going to use the method of complements. We can identify this probability by taking 1 and subtracting the probability of the complement, meaning x must be less than 1, and the only value that we can see is X equals 0, right? So basically, the probability of X being greater than or equal to 1 is equal to 1 minus the probability of X being 0. For this problem we know that the probability of a single failuree is equal to 0.074, so we want to identify the complemente bar which indicates the probability of a success, right, which is 1 minus p and that's 1 minus 0.074. So we have defined the probability of a single success. Now P X equals 0 represents the probability of 5 successes. So what we want to do is basically apply the multiplication rule for independent events. We're going to take 1 minus 0.074, which is the probability of a single success and raise it to the power of 5, right, using the multiplication rule. And now substituting it back we get E of X being greater than or equal to 1. I 1 minus. 1 minus 0.074 raises the power of 5. When we perform the calculations, we're going to get 0.319. So that is our first answer, and now we want to answer the second question. Should the company prepare for frequent individual rechecks? The answer is yes. Our probability is significantly higher, right? 0.319 is roughly 1/3, so this is quite a high probability. Thank you for watching.

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Key Concepts

Probability of Events

Binomial Distribution

Complement Rule

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