Problem
In a study about a pandemic, data of 900 persons was collected. It was found that:
- 190 persons had symptom of fever
- 220 persons had symptom of cough
- 220 persons had symptom of breathing problem
- 330 persons had symptom of fever or cough or both
- 350 persons had symptom of cough or breathing problem or both
- 340 persons had symptom of fever or breathing problem or both
- 30 persons had all three symptoms (fever, cough, and breathing problem)
If a person is chosen randomly from these 900 persons, then find the probability that the person has at most one symptom.
Solution
To find the number of people that have atmost one symptom, we can add the number of people that have no symptoms and those that have exactly one symptom.
Or, we can subtract the number of people that have two or more symptoms from the total number of people. I’ll discuss the second method.
Let F, C, and B respectively denote the set of people that have fever, cough, and breathing problem.
Using the first three statements, we get n(F) = 190, n(C) = 220, n(B) = 220.
Using the next three statements, we get n(F U C) = 330, n(C U B) = 350, n(B U F) = 340
Finally, using the last statement, we get n(F ∩ C ∩ B) = 30.
Now, the number of people that have two or more symptoms can be expressed as the sum of people that have exactly two symptoms and the number of people that have all three symptoms.
The latter has been given directly, which is n(F ∩ C ∩ B) or 30.
To find the former, let’s write out the expression representing it.
n(F ∩ C) represents the number of people that have fever and cough both. But they may have breathing problems as well.
So, to find the number of people that have exactly the two symptoms, fever and cough, we’ll remove the number of people that have breathing problem as well, i.e. n(F ∩ C ∩ B).
So, the number of people that have only cough and fever (but not breathing problem) is n(F ∩ C) – n(F ∩ C ∩ B). By the way, this is nothing but n(F ∩ C ∩ B’).
Similarly, the number of people that have only cough and breathing problem (but not fever) is n(C ∩ B) – n(F ∩ C ∩ B).
Again, the number of people that have only breathing problem and fever (but not cough) is n(B ∩ F) – n(F ∩ C ∩ B).
Finally, the number of people that have exactly two symptoms is the sum of the above three expressions, which is n(F ∩ C) + n(C ∩ B) + n(B ∩ F) – 3n(F ∩ C ∩ B).
And, if we add n(F ∩ C ∩ B) to the above, we get the number of people that have two or more symptoms. We’ll get the expression n(F ∩ C) + n(C ∩ B) + n(B ∩ F) – 2n(F ∩ C ∩ B).
Now, we aren’t given the values of n(F ∩ C), n(C ∩ B), or n(B ∩ F).
But using n(F U C) = n(F) + n(C) – n(F ∩ C), we have n(F ∩ C) = n(F) + n(C) – n(F U C).
Using similar relations for the other two, we get the number of people with two or more symptoms as 2n(F) + 2n(C) + 2n(B) – n(F U C) – n(C U B) – n(B U F) – 2n(F ∩ C ∩ B).
Now, let’s substitute the values to find this number – we’ll get 2(190) + 2(220) + 2(220) – 330 – 350 – 340 – 2(30), which simplifies to 180.
Finally, the number of people that have atmost one symptom will be 900 – 180 or 720, which means the required probability equals 720/900, which simplifies to 4/5 or 0.8.
Comments
A straightforward problem, typically found in standard textbooks. Not quite deserving of the JEE advanced.