Probability of Random Person Having At Most One Symptom in Pandemic Study

To find the probability that a randomly chosen person has at most one symptom, we can calculate the number of persons having at most one symptom using set theory principles.

Let $F$, $C$, and $B$ denote the sets of persons having symptoms of fever, cough, and breathing problem, respectively.

From the given data, the total number of persons is $N = 900$.
We are given:
$n(F) = 190$
$n(C) = 220$
$n(B) = 220$

$n(F \cup C) = 330$
$n(C \cup B) = 350$
$n(F \cup B) = 340$

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

Using the principle of inclusion-exclusion for two sets:
$n(F \cup C) = n(F) + n(C) - n(F \cap C)$
$330 = 190 + 220 - n(F \cap C) \implies n(F \cap C) = 410 - 330 = 80$

$n(C \cup B) = n(C) + n(B) - n(C \cap B)$
$350 = 220 + 220 - n(C \cap B) \implies n(C \cap B) = 440 - 350 = 90$

$n(F \cup B) = n(F) + n(B) - n(F \cap B)$
$340 = 190 + 220 - n(F \cap B) \implies n(F \cap B) = 410 - 340 = 70$

Now, we calculate the number of persons who have at least two symptoms:
The number of persons having exactly two symptoms is given by:
$n_2 = [n(F \cap C) - n(F \cap C \cap B)] + [n(C \cap B) - n(F \cap C \cap B)] + [n(F \cap B) - n(F \cap C \cap B)]$
$n_2 = (80 - 30) + (90 - 30) + (70 - 30) = 50 + 60 + 40 = 150$

The number of persons having all three symptoms is:
$n_3 = n(F \cap C \cap B) = 30$

Therefore, the total number of persons having at least two symptoms is:
$n(\text{at least 2 symptoms}) = n_2 + n_3 = 150 + 30 = 180$

The number of persons having at most one symptom is:
$n(\text{at most 1 symptom}) = N - n(\text{at least 2 symptoms}) = 900 - 180 = 720$

The probability $P$ that a randomly chosen person has at most one symptom is:
$P = \frac{n(\text{at most 1 symptom})}{N} = \frac{720}{900} = 0.8$