Question

scientists have found a way to distinguish chronic - fatigue syndrome (cfs) from post - treatment lyme disease by different proteins in a patients spinal fluid. in a study of 4105 patients, the results to the right were found. (a) how many had the protein for lyme only? (b) how many had the normal protein only? (c) how many had none of the proteins? 1609 had the protein for normal patients, cfs, and post - treatment lyme disease; 1898 had the protein for cfs and lyme; 621 had the protein for cfs only; 2792 had the protein for cfs; 1749 had the normal protein as well as the one for lyme; 2725 had the protein for lyme; 2508 had the normal protein. (a) there were patients who only had the protein for lyme. (type a whole number.)

Explanation:

Step1: Find number of Lyme - only patients

We know the total number of patients with Lyme protein is 2725 and the number of patients with both CFS and Lyme protein is 1898. The number of patients with only Lyme protein is the total Lyme - positive patients minus those with both CFS and Lyme.
$2725 - 1898=827$

Step2: Find number of normal - only patients

The number of patients with normal protein is 2508 and the number of patients with normal and Lyme protein is 1749 and the number of patients with normal, CFS and post - treatment Lyme protein is 1609. First, find the number of patients with normal protein and other proteins. Let \(N\) be the number of patients with normal protein and other proteins. \(N = 1749+1609\). Then the number of normal - only patients is \(2508-(1749 + 1609 - 1609)=2508 - 1749=759\)

Step3: Find number of patients with none of the proteins

Let \(A\) be the set of CFS - positive patients, \(B\) be the set of Lyme - positive patients and \(C\) be the set of normal - protein patients.
The number of patients with at least one of the proteins:
The number of CFS - positive patients \(n(A)=2792\), Lyme - positive patients \(n(B)=2725\) and normal - protein patients \(n(C)=2508\). The number of patients with both CFS and Lyme \(n(A\cap B)=1898\), normal and Lyme \(n(C\cap B)=1749\), normal, CFS and post - treatment Lyme \(n(C\cap A\cap B) = 1609\), CFS only \(n(A\cap\overline{B}\cap\overline{C})=621\)
Using the principle of inclusion - exclusion:
\[n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(A\cap C)-n(B\cap C)+n(A\cap B\cap C)\]
We know \(n(A\cap C)\) is part of the 1609 and other combinations.
The number of patients with at least one of the proteins:
\[n(A\cup B\cup C)=2792 + 2725+2508-1898 - 1749+1609\]
\[n(A\cup B\cup C)=2792+2725 + 2508-(1898 + 1749)+1609\]
\[n(A\cup B\cup C)=2792+2725+2508 - 3647+1609\]
\[n(A\cup B\cup C)=6087\]
The number of patients with none of the proteins is \(4105 - 6087+1609=277\)

Answer:

(a) 827
(b) 759
(c) 277