In a survey of 100 patients who reported at a hospital one day, it was found out yhat 70 complained of fever, 50 complained of stomach ache and 30 were injured. All 100 patients had at least onr of the complaints and 44 had exactly two of the complaints. How many patients had all three complaints?
Let's use the principle of inclusion-exclusion to solve this problem.
Let's define:
F = number of patients with a fever
S = number of patients with a stomach ache
I = number of patients with an injury
FSI = number of patients with all three complaints: fever, stomach ache, and injury
FS, FI, SI = number of patients with exactly two of the complaints
According to the problem, we have:
F = 70
S = 50
I = 30
FS + FI + SI - 3*FSI = 44 (because each patient with all three complaints is counted three times in the sum of the pairs, we need to subtract three times that amount to correct the count)
FS + FI + SI = 44 + 3*FSI
Now, according to the principle of inclusion-exclusion, the total number of patients, which is 100, is equal to the sum of the individual complaints minus the sum of the pairs plus the number of people with all three complaints:
F + S + I - (FS + FI + SI) + FSI = 100
Plugging in the values we have:
70 + 50 + 30 - (44 + 3*FSI) + FSI = 100
150 - 44 - 3*FSI + FSI = 100
106 - 2*FSI = 100
2*FSI = 106 - 100
2*FSI = 6
FSI = 6 / 2
FSI = 3
We found that there are 3 patients who had all three complaints: fever, stomach ache, and injury.