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Probability question

2022 · Shift 1 · Q21
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Probability question

2022 · Shift 1 · Q21

JEE AdvancedMathematicsProbabilityNumerical+3 / −1
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 ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 0.79TO0.81

Let

  • FFF = set of persons having fever,
  • CCC = set of persons having cough,
  • BBB = set of persons having breathing problem.

Given:

\quad |C|=220, \quad |B|=220$$ $$|F\cup C|=330, \quad |C\cup B|=350, \quad |F\cup B|=340$$ $$|F\cap C\cap B|=30$$ Total persons $=900$. We need the probability that a randomly chosen person has **at most one symptom**, i.e. either: 1. no symptom, or 2. exactly one symptom. --- ## 1. Find pairwise intersections Using $$|A\cup B|=|A|+|B|-|A\cap B|$$ ### (i) Fever and cough $$330=190+220-|F\cap C|$$ $$|F\cap C|=410-330=80$$ ### (ii) Cough and breathing problem $$350=220+220-|C\cap B|$$ $$|C\cap B|=440-350=90$$ ### (iii) Fever and breathing problem $$340=190+220-|F\cap B|$$ $$|F\cap B|=410-340=70$$ So, $$|F\cap C|=80, \quad |C\cap B|=90, \quad |F\cap B|=70$$ --- ## 2. Find number of persons having at least one symptom By inclusion-exclusion: $$|F\cup C\cup B|=|F|+|C|+|B|-|F\cap C|-|C\cap B|-|F\cap B|+|F\cap C\cap B|$$ Substitute values: $$|F\cup C\cup B|=190+220+220-80-90-70+30$$ $$=630-240+30=420$$ Thus, number of persons with **at least one symptom** is $$420$$ Hence, number of persons with **no symptom** is $$900-420=480$$ --- ## 3. Find number of persons having exactly one symptom First find those having exactly two symptoms. Since the pairwise intersections include the triple intersection, we get: $$\text{Exactly }(F\cap C)=80-30=50$$ $$\text{Exactly }(C\cap B)=90-30=60$$ $$\text{Exactly }(F\cap B)=70-30=40$$ So number having exactly two symptoms is $$50+60+40=150$$ Number having exactly three symptoms is $$30$$ Therefore, number having exactly one symptom is $$420-150-30=240$$ --- ## 4. Find number having at most one symptom At most one symptom = no symptom + exactly one symptom $$=480+240=720$$ Therefore required probability is $$\frac{720}{900}=\frac{4}{5}=0.8$$ --- ## 5. Final Answer $$\boxed{0.8}$$ Since this is an integer/numerical-type response in decimal form, the answer is $0.8$ (equivalently $\frac45$). This lies in the stored range $0.79$ to $0.81$.
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