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 ______.

Let the sets be: $$F$$ = persons with fever, $$C$$ = persons with cough, $$B$$ = persons with breathing problem.

The given data are:
$$|F| = 190,\; |C| = 220,\; |B| = 220$$
$$|F \cup C| = 330,\; |C \cup B| = 350,\; |F \cup B| = 340$$
$$|F \cap C \cap B| = 30$$

Denote the pair-wise intersections (each includes the triple intersection) by
$$x = |F \cap C|,\; y = |C \cap B|,\; z = |F \cap B|.$$ Using the union formula $$|A \cup B| = |A| + |B| - |A \cap B|$$ we get:

For $$F$$ and $$C$$:
$$190 + 220 - x = 330 \;\Rightarrow\; x = 80.$$

For $$C$$ and $$B$$:
$$220 + 220 - y = 350 \;\Rightarrow\; y = 90.$$

For $$F$$ and $$B$$:
$$190 + 220 - z = 340 \;\Rightarrow\; z = 70.$$

Split every region of the Venn diagram:
Case 1: exactly two symptoms (excluding the third):
$$|F \cap C \text{ only}| = 80 - 30 = 50,$$
$$|C \cap B \text{ only}| = 90 - 30 = 60,$$
$$|F \cap B \text{ only}| = 70 - 30 = 40.$$

Case 2: exactly one symptom. Let these counts be $$a,b,c$$ respectively:

From $$|F| = a + 50 + 40 + 30$$ $$190 = a + 120 \;\Rightarrow\; a = 70.$$ From $$|C| = b + 50 + 60 + 30$$ $$220 = b + 140 \;\Rightarrow\; b = 80.$$ From $$|B| = c + 60 + 40 + 30$$ $$220 = c + 130 \;\Rightarrow\; c = 90.$$

Case 3: all three symptoms: $$g = 30.$$

Total counted in Venn regions:
$$70 + 80 + 90 + 50 + 60 + 40 + 30 = 420.$$

Hence the number having none of the symptoms is
$$h = 900 - 420 = 480.$$

People with at most one symptom = (exactly one) + (none)
$$= (70 + 80 + 90) + 480 = 240 + 480 = 720.$$

Required probability $$= \frac{720}{900} = 0.80.$$

Therefore, the probability that a randomly chosen person has at most one symptom is 0.80.