Question 3 - Jee advanced Math 2022 P1 Questions with Solutions

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 _____.

Sol : 

Let C be a letter for representing ‘Cough’; F for ‘Fever’; and B for ‘breathing problem’. 

i) 190 = n(F) = n(F only) + n(F and C) + n(F and B) + n(all three)

ii) 220 = n(C) = n(C only) + n(F and C) + n(C and B) + n(all three)

iii) 220 = n(B) = n(B only) + n(F and B) + n(C and B) + n(all three)

n(F or C or both) = n(F) + n(C) - n(F and C) - n(all three)
(We are subtracting n(F and C) because it is already included in both n(F) and n(C) values(see 1 and 2 above). So we need to subtract one of them to avoid repetition. Same reason for subtracting n(all three).)

\(\implies\) 330 = 190 + 220 - n(F and C) - 30

\(\implies\)n(F and C) = 50

Similarly, n(C or B or both) = n(C) + n(B) - n(C and B) - n(all three)

\(\implies\) 350 = 220 + 220 - n(C and B) - 30

\(\implies\) n(C and B) = 60

And, n(F or B or both) = n(F) + n(B) - n(F and B) - n(all three)

\(\implies\) 340 = 190 + 220 - n(F and B) - 30

\(\implies\) n(F and B) = 40

From(i),
190 = n(F) = n(F only) + n(F and C) + n(F and B) + n(all three)
\(\implies\) n(F only) = 190 - 50 - 40 - 30
                        = 70

From (ii),
220 = n(C) = n(C only) + n(F and C) + n(C and B) + n(all three)
\(\implies\) n(C only) = 220 - 50 - 60 - 30
                        = 80

From (iii),
220 = n(B) = n(B only) + n(F and B) + n(C and B) + n(all three)
\(\implies\) n(B only) = 220 - 40 - 60 - 30
                        = 90

Total persons = 900

900 = n(all three symptoms) + n(no symptoms) + n(F only) + n(C only) + n(B only) + n(F and C) + n(C and B) + n(F and B)

\(\implies\) n(no symptoms) = 900 - (30 + 70 + 80 + 90 + 50 + 60 + 40)
                     = 480

P(at-most one symptom) = P(only one symptom) + P(no symptoms)
                                        = (P(F only) + P(C only) + P(B only)) + P(no symptoms)
                                        = \(\frac{70 + 80 + 90 + 480}{900}\)
                                        = \(\frac{8}{10}\)
                                         = 0.8

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