Suppose one of the lab assistants accidentally mixed two patients' blood samples before they were distributed to the vials. Which of the following correctly represents the set of all possible numbers of positive test results out of the eight vials?

Correct Answer: The correct option is {4,5,6,7,8}.

Step-by-Step Explanation:

To understand why this set represents all possible numbers of positive test results, let us break down the testing structure and the logical consequences of the lab assistant's error.

1. Understanding the Setup
In this problem, we have 16 patients and 8 vials (labeled A through H). Exactly one of the patients is infected with the disease. The blood samples are pooled such that each patient's blood sample is distributed into exactly 4 vials.
Under normal conditions, since the infected patient's blood is in exactly 4 vials, those 4 vials will test positive, and the remaining 4 vials will test negative. Thus, with no mix-up, there are exactly 4 positive vials.

2. Analyzing the Accidental Mixing
Suppose the lab assistant accidentally mixes the blood samples of two distinct patients, say Patient X and Patient Y. Let V_X be the set of 4 vials assigned to Patient X, and V_Y be the set of 4 vials assigned to Patient Y. Since each patient is assigned exactly 4 vials, we have:

$\left|V_X\right|=4$ and $\left|V_Y\right|=4$

We analyze the two possible scenarios based on which patient is infected:

Case 1: The infected patient is neither Patient X nor Patient Y
If the infected patient is a third patient (say, Patient Z), then the mixed sample of X and Y does not contain the disease. Patient Z's sample is distributed normally to its 4 assigned vials. Therefore, only the 4 vials containing Patient Z's blood will test positive. This gives us 4 positive test results.

Case 2: The infected patient is either Patient X or Patient Y
Without loss of generality, let Patient X be the infected patient. Since Patient X's blood is mixed with Patient Y's blood, the combined mixed sample now contains the disease. Since this mixed sample is distributed to all the vials in V_X and V_Y, every vial that receives blood from either Patient X or Patient Y will test positive.
The total number of positive vials is therefore the cardinality of the union of their assigned vials:

$|V_X\cup V_Y|$

By the Principle of Inclusion-Exclusion, the number of vials in the union is calculated as:

$|V_X\cup V_Y|=\left|V_X\right|+\left|V_Y\right|-|V_X\cap V_Y|$

Substituting the values of |V_X| and |V_Y|:

$|V_X\cup V_Y|=4+4-|V_X\cap V_Y|=8-|V_X\cap V_Y|$

Since Patient X and Patient Y are distinct patients, their assigned vial sets must be different, which means they cannot share all 4 vials. Therefore, the intersection size

$|V_X\cap V_Y|$

can be any integer from 0 to 3. Let's calculate the corresponding number of positive test results for each possible intersection size:

• If they share 3 vials: 8 - 3 = 5 positive test results.
• If they share 2 vials: 8 - 2 = 6 positive test results.
• If they share 1 vial: 8 - 1 = 7 positive test results.
• If they share 0 vials: 8 - 0 = 8 positive test results.

Conclusion
Combining the results from Case 1 and Case 2, the possible numbers of positive test results out of the eight vials are 4, 5, 6, 7, and 8. Thus, the set of all possible positive test results is {4, 5, 6, 7, 8}.