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JEE Advanced 2022 · Paper 1, Mathematics question 3

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 _____.

Answer and worked solution

Official answer: 0.79

Let F, C, B be the three symptom sets.

Pairwise intersections (from ∣X∪Y∣=∣X∣+∣Y∣−∣X∩Y∣|X\cup Y|=|X|+|Y|-|X\cap Y|):

  • ∣F∩C∣=190+220−330=80|F\cap C|=190+220-330=80;
  • ∣C∩B∣=220+220−350=90|C\cap B|=220+220-350=90;
  • ∣F∩B∣=190+220−340=70|F\cap B|=190+220-340=70.
    Also ∣F∩C∩B∣=30|F\cap C\cap B|=30.

Union.
∣F∪C∪B∣=630−240+30=420,|F\cup C\cup B|=630-240+30=420,
so 900−420=480900-420=480 people have no symptom.

Exactly one symptom.
∑∣X∣−2∑∣X∩Y∣+3∣F∩C∩B∣=630−480+90=240.\sum|X|-2\sum|X\cap Y|+3|F\cap C\cap B|=630-480+90=240.

Answer.
P(at most one)=480+240900=0.8.P(\text{at most one})=\frac{480+240}{900}=0.8.

Topics:Classical probabilitySets and operations
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