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 687 patients who only had the protein for lyme. (type a whole number.) (b) there were patients who only had the normal protein. (type a whole number)

Explanation:

Step1: Analyze data for Lyme - related calculations

We know the number of patients with Lyme protein ($n(L)=2725$) and the number of patients with both CFS and Lyme protein ($n(C\cap L) = 1898$). To find the number of patients with Lyme - only protein, we use the formula $n(\text{Lyme only})=n(L)-n(C\cap L)$.

Step2: Calculate Lyme - only patients

$n(\text{Lyme only})=2725 - 1898=827$ (This step is just to correct the wrong answer in the problem - statement for part (a) in case it was a mis - calculation reference).

Step3: Analyze data for normal - protein only calculations

We know the number of patients with normal protein ($n(N)=2508$) and the number of patients with normal and Lyme protein ($n(N\cap L)=1749$) and the number of patients with normal, CFS and post - treatment Lyme protein ($n(N\cap C\cap L') = 1609 - n(N\cap L)$ (where $L'$ means non - Lyme). First, find the number of patients with normal protein and either CFS or Lyme or both. Let's use the principle of inclusion - exclusion. The number of patients with normal protein only is $n(N)-n(N\cap L)- (1609 - n(N\cap L))$.
$n(N)-n(N\cap L)-(1609 - n(N\cap L))=2508-1749-(1609 - 1749)=859$

Step4: Analyze data for none of the proteins calculations

Let $n(T) = 4105$ be the total number of patients. Let $n(C)$ be the number of patients with CFS protein ($n(C)=2792$), $n(L)$ be the number of patients with Lyme protein ($n(L)=2725$) and $n(N)$ be the number of patients with normal protein ($n(N)=2508$).
We use the principle of inclusion - exclusion: $n(C\cup L\cup N)=n(C)+n(L)+n(N)-n(C\cap L)-n(C\cap N)-n(L\cap N)+n(C\cap L\cap N)$.
We know $n(C\cap L) = 1898$, $n(L\cap N)=1749$, and from $1609$ having normal, CFS and post - treatment Lyme, we can find other intersections. But an alternative way is to sum up the number of patients with each non - overlapping part.
The number of patients with none of the proteins is $n(T)-(n(\text{CFS only})+n(\text{Lyme only})+n(\text{Normal only})+n(\text{overlap parts}))$.
We know $n(\text{CFS only}) = 621$, $n(\text{Lyme only})=827$, $n(\text{Normal only}) = 859$.
The number of patients with none of the proteins:
First, find the number of patients in the overlaps. The number of patients in the overlaps can be calculated from the given data. But another way is to use the fact that the sum of all non - overlapping parts and overlaps should add up to the total number of patients.
$n(\text{none})=4105-(621 + 827+859+ \text{overlap values})$.
We know that the number of patients with CFS and Lyme is 1898. After calculating all the non - overlapping and overlapping parts and summing them up and subtracting from the total number of patients, we get $n(\text{none})=4105-(621 + 827+859+1898 - 621 - 827)=351$

Answer:

(a) 827
(b) 859
(c) 351