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QJEEAdv22Mac03a: JEE Advanced 2022 Mathematics - Class 12 JEEAdv22Mac03a Solution: In a study about a pandemic, data of 900 persons was collected. …

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 ____________.

Step-by-Step Verified Solution

Let person had symptom of fever =n(F)=190=n(F)=190

Person had symptom of cough =n(C)=220=n(\mathrm{C})=220

Person had symptom of breathing =n( B)=220=n(\mathrm{~B})=220

Person had symptom of fever or cough

=n( F∪C)=330

Person had symptom of cough or breathing

=n(C∪B)=350

Person had symptom of fever or breathing

=n( F∪B)=340

Person had symptom of fever, cough and breathing =n( F∩C∪B)=30=n(\mathrm{~F} \cap \mathrm{C} \cup \mathrm{B})=30

So n( F∩C)=n( F)+n(C)−n( F∪C)n(\mathrm{~F} \cap \mathrm{C})=n(\mathrm{~F})+n(\mathrm{C})-n(\mathrm{~F} \cup \mathrm{C})

=190+220−330=190+220-330

=80=80

and

n(F∩B)=n( F)+n( B)−n( F∪B)n(\mathrm{F} \cap \mathrm{B}) =n(\mathrm{~F})+n(\mathrm{~B})-n(\mathrm{~F} \cup \mathrm{B})

=190+220−340=190+220-340

=70=70

and n(C∩B)=n(C)+n( B)−n(C∩B)n(\mathrm{C} \cap \mathrm{B}) =n(\mathrm{C})+n(\mathrm{~B})-n(\mathrm{C} \cap \mathrm{B})

=220+220−350=220+220-350

=90=90

Number of people having at most one symptom

=70+80+90+480=720

JEE Advanced 2022 Paper 1 Online Mathematics - Probability Question 12 English Explanation

Required probability =720900=0.80=\frac{720}{900}=0.80

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