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AP Statistics 2.5 Mutually Exclusive Events- Exam Style Questions - FRQs - New Syllabus

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Question

The ELISA tests whether a patient has contracted HIV. The ELISA is said to be positive if it indicates that HIV is present in a blood sample, and the ELISA is said to be negative if it does not indicate that HIV is present in a blood sample. Instead of directly measuring the presence of HIV, the ELISA measures levels of antibodies in the blood that should be elevated if HIV is present. Because of variability in antibody levels among human patients, the ELISA does not always indicate the correct result.
As part of a training program, staff at a testing lab applied the ELISA to 500 blood samples known to contain HIV. The ELISA was positive for 489 of those blood samples and negative for the other 11 samples. As part of the same training program, the staff also applied the ELISA to 500 other blood samples known to not contain HIV. The ELISA was positive for 37 of those blood samples and negative for the other 463 samples.
(a) When a new blood sample arrives at the lab, it will be tested to determine whether HIV is present. Using the data from the training program, estimate the probability that the ELISA would be positive when it is applied to a blood sample that does not contain HIV.
(b) Among the blood samples examined in the training program that provided positive ELISA results for HIV, what proportion actually contained HIV?
(c) When a blood sample yields a positive ELISA result, two more ELISAs are performed on the same blood sample. If at least one of the two additional ELISAs is positive, the blood sample is subjected to a more expensive and more accurate test to make a definitive determination of whether HIV is present in the sample. Repeated ELISAs on the same sample are generally assumed to be independent. Under the assumption of independence, what is the probability that a new blood sample that comes into the lab will be subjected to the more expensive test if that sample does not contain HIV?

Most-appropriate topic codes (AP Statistics):

• Topic \(2.6\) — Conditional Probability (Parts \(\mathrm{a}\), \(\mathrm{b}\))
• Topic \(2.7\) — Independent Events and Unions of Events (Part \(\mathrm{c}\))
• Topic \(2.5\) — Mutually Exclusive Events (Part \(\mathrm{c}\), complement rule)
▶️ Answer/Explanation

(a)

Out of the 500 blood samples known to not contain HIV, the ELISA returned a positive result for 37 of them.
So the estimated probability that the ELISA is positive given no HIV is:
\(P(\text{positive} \mid \text{no HIV}) = \frac{37}{500} = 0.074\)
\(\boxed{P(\text{positive} \mid \text{no HIV}) \approx 0.074}\)

(b)

First, find the total number of blood samples that returned a positive ELISA result across both groups:
\(489 + 37 = 526 \text{ positive results in total}\)
Of those 526 positive results, 489 actually came from samples that truly contained HIV.
So the proportion of positive ELISA results that actually contained HIV is:
\(\frac{489}{526} \approx 0.9297\)
\(\boxed{\frac{489}{526} \approx 0.9297}\)

(c)

From part (a), the probability the ELISA is positive for a sample with no HIV is \(0.074\), and therefore the probability it is negative is \(1 – 0.074 = 0.926\).
A sample with no HIV will be sent for the expensive test if: the 1st ELISA is positive AND at least one of the 2 follow-up ELISAs is also positive.
Find the probability that at least one of the two follow-up tests is positive (using the complement):
\(P(\text{at least one positive in 2 follow-ups}) = 1 – P(\text{both negative}) = 1 – (0.926)^2\)
\(= 1 – 0.857476 = 0.142524\)
Now multiply by the probability the first ELISA is positive:
\(P(\text{subjected to expensive test}) = (0.074)(0.142524)\)
\(= 0.010547\)
\(\boxed{P(\text{subjected to expensive test}) \approx 0.0105}\)

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