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 486 patients who only had the normal protein. (type a whole number.) (c) there were blank patients who had none of the proteins. (type a whole number.)

Explanation:

Step1: Calculate the number of patients with none of the proteins

Let \(N = 4105\) be the total number of patients.
Let \(A\) be the set of patients with CFS - related protein, \(|A|=2792\), \(B\) be the set of patients with Lyme - related protein, \(|B| = 2725\), and \(C\) be the set of patients with normal protein, \(|C|=2508\).
We know that \(|A\cap B| = 1898\), \(|A\cap C|=1609\), \(|B\cap C| = 1749\).
The number of patients with all three proteins is \(|A\cap B\cap C|\).
We know that \(|A|=|A\cap B\cap C|+(|A\cap B|-|A\cap B\cap C|)+(|A\cap C|-|A\cap B\cap C|)+ \text{(patients with only }A)\). So \(|A\cap B\cap C|=|A|- \text{(patients with only }A)-( |A\cap B|-|A\cap B\cap C|)-( |A\cap C|-|A\cap B\cap C|)\). Given \(|A| = 2792\) and \(\text{(patients with only }A)=621\), \(|A\cap B| = 1898\), \(|A\cap C|=1609\).
First, find \(|A\cap B\cap C|\):
We know that \(|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|-\text{(patients with only }A)-\text{(patients with only }B)-\text{(patients with only }C)\) is part of the inclusion - exclusion principle.
We already know \(\text{(patients with only }A) = 621\), \(\text{(patients with only }B)\) (from part (a) is \(687\)), \(\text{(patients with only }C)\) (from part (b) is \(486\)).
By the inclusion - exclusion principle \(|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|\).
We first find \(|A\cap B\cap C|\):
We know \(|A| = 2792\), \(|B| = 2725\), \(|C| = 2508\), \(|A\cap B| = 1898\), \(|A\cap C|=1609\), \(|B\cap C| = 1749\).
\[

$$\begin{align*} |A\cup B\cup C|&=2792 + 2725+2508-1898 - 1609-1749+|A\cap B\cap C|\\ \end{align*}$$

\]
We can also calculate it in another way.
The number of patients with at least one of the proteins:
\[

$$\begin{align*} &621+687 + 486+1898+1609+1749 - 2\times|A\cap B\cap C|\\ \end{align*}$$

\]
We know that \(|A\cap B| = 1898\), \(|A| = 2792\), so the number of patients in \(|A\cap B\) but not only in \(|A\cap B|\) is \(1898 - |A\cap B\cap C|\). Similarly for \(|A\cap C|\) and \(|B\cap C|\).
The number of patients with at least one of the proteins is \(621+687+486+(1898 - |A\cap B\cap C|)+(1609 - |A\cap B\cap C|)+(1749 - |A\cap B\cap C|)+|A\cap B\cap C|\)
\[

$$\begin{align*} &621+687+486+1898+1609+1749-2|A\cap B\cap C|\\ &=(621 + 687+486)+(1898+1609+1749)-2|A\cap B\cap C|\\ &=1794+5256-2|A\cap B\cap C| \end{align*}$$

\]
We use the inclusion - exclusion principle \(|A\cup B\cup C|=|A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|\)
\[

$$\begin{align*} |A\cup B\cup C|&=2792+2725 + 2508-1898-1609-1749+|A\cap B\cap C|\\ &=3779+|A\cap B\cap C| \end{align*}$$

\]
Equating the two expressions for \(|A\cup B\cup C|\):
\[

$$\begin{align*} 1794 + 5256-2|A\cap B\cap C|&=3779+|A\cap B\cap C|\\ 3|A\cap B\cap C|&=1794 + 5256-3779\\ 3|A\cap B\cap C|&=3271\\ |A\cap B\cap C|&= 1090 \end{align*}$$

\]
The number of patients with at least one of the proteins is \(|A\cup B\cup C|=2792+2725+2508-1898-1609-1749 + 1090=3779\)
The number of patients with none of the proteins is \(N-|A\cup B\cup C|=4105 - 3779=326\)

Answer:

326