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 4064 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? 1655 had the protein for normal patients, cfs, and post - treatment lyme disease; 1855 had the protein for cfs and lyme; 647 had the protein for cfs only; 2642 had the protein for cfs; 1844 had the normal protein as well as the one for lyme; 2693 had the protein for lyme; 2568 had the normal protein. (a) there were patients who only had the protein for lyme. (type a whole number.)

Explanation:

Step1: Find number of patients with both CFS and Lyme

We know that 2642 had the protein for CFS and 647 had the protein for CFS only. So the number of patients with both CFS and Lyme is $2642 - 647=1995$.

Step2: Find number of patients with Lyme only

We know that 2693 had the protein for Lyme. The number of patients with both CFS and Lyme is 1995. So the number of patients with Lyme only is $2693 - 1995 = 698$.

Step3: Find number of patients with normal protein only

We know that 1655 had the protein for normal patients, CFS and post - treatment Lyme disease, and 1844 had the normal protein as well as the one for Lyme. Also, 2568 had the normal protein. Let \(N\) be the number of patients with normal protein only.
First, find the number of patients with normal protein and either CFS or Lyme or both. Let \(A\) be the set of normal - protein patients, \(B\) be the set of CFS - protein patients and \(C\) be the set of Lyme - protein patients.
We know \(|A\cap B\cap C| = 1655\), \(|A\cap C|=1844\), \(|A| = 2568\).
The number of patients with normal protein and either CFS or Lyme or both is \(|(A\cap B)\cup(A\cap C)|=|A\cap B|+|A\cap C|-|A\cap B\cap C|\). Since \(|A\cap B|+|A\cap C|-|A\cap B\cap C|\) is part of the normal - protein patients, the number of normal - protein only patients is \(2568-(1844 + 1655-1655)=724\).

Step4: Find number of patients with none of the proteins

Let \(U = 4064\) (total number of patients).
The number of patients with at least one of the proteins:
Number of CFS - related patients \(n(CFS)=2642\), number of Lyme - related patients \(n(Lyme)=2693\), number of normal - protein patients \(n(Normal)=2568\).
Using the principle of inclusion - exclusion \(n(CFS\cup Lyme\cup Normal)=n(CFS)+n(Lyme)+n(Normal)-n(CFS\cap Lyme)-n(CFS\cap Normal)-n(Lyme\cap Normal)+n(CFS\cap Lyme\cap Normal)\).
We know \(n(CFS\cap Lyme) = 1995\), \(n(CFS\cap Normal)\) is part of 1655, \(n(Lyme\cap Normal)=1844\), \(n(CFS\cap Lyme\cap Normal)=1655\).
\(n(CFS\cup Lyme\cup Normal)=2642 + 2693+2568-1995 - 1655-1844 + 1655=3064\).
The number of patients with none of the proteins is \(4064 - 3064=1000\).

Answer:

(a) 698
(b) 724
(c) 1000