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 the probability that the person has at most one symptom is ______________.

The correct answer is 0.8 (or 0.80).

Let us analyze the given problem step-by-step using set theory and the Principle of Inclusion-Exclusion.

1. Given Information:
Total number of persons surveyed, $N=900$.
Let $F$ be the set of persons having fever.
Let $C$ be the set of persons having cough.
Let $B$ be the set of persons having breathing problem.

From the given data, we have:

$n\left(F\right)=190$

$n\left(C\right)=220$

$n\left(B\right)=220$

$n(F\cup C)=330$

$n(C\cup B)=350$

$n(F\cup B)=340$

$n(F\cap C\cap B)=30$

2. Finding Pairwise Intersections:
Using the formula for the union of two sets, $n(X\cup Y)=n\left(X\right)+n\left(Y\right)-n(X\cap Y)$:

For Fever and Cough:

$n(F\cap C)=n\left(F\right)+n\left(C\right)-n(F\cup C)=190+220-330=80$

For Cough and Breathing Problem:

$n(C\cap B)=n\left(C\right)+n\left(B\right)-n(C\cup B)=220+220-350=90$

For Fever and Breathing Problem:

$n(F\cap B)=n\left(F\right)+n\left(B\right)-n(F\cup B)=190+220-340=70$

3. Finding the Number of Persons with at Most One Symptom:
"At most one symptom" means a person has either 0 symptoms or exactly 1 symptom.
Equivalently, it is equal to the total number of persons minus the persons having 2 or 3 symptoms (i.e., at least 2 symptoms).

Let us calculate the number of persons having at least 2 symptoms:

Number of persons having exactly 2 symptoms is given by:

$\left[n\right(F\cap C)-n(F\cap C\cap B\left)\right]+\left[n\right(C\cap B)-n(F\cap C\cap B\left)\right]+\left[n\right(F\cap B)-n(F\cap C\cap B\left)\right]$

$=(80-30)+(90-30)+(70-30)=50+60+40=150$

Number of persons having all 3 symptoms = 30.

Therefore, the total number of persons with at least 2 symptoms:

$\text{At least 2 symptoms}=150+30=180$

Hence, the number of persons with at most one symptom:

$\text{At most 1 symptom}=\text{Total persons}-\text{Persons with at least 2 symptoms}$

$=900-180=720$

4. Calculating the Required Probability:
The probability that a randomly chosen person has at most one symptom is:

$P\left(\text{at most 1 symptom}\right)=\frac{720}{900}=0.80$

Thus, the required probability is 0.8 (or 0.80).