In a study about a pandemic,data of 900 perso | vedclass

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(A) $n(U) = 900$Let $A \equiv 	ext{Fever}$,$B \equiv 	ext{Cough}$,$C \equiv 	ext{Breathing problem}$.Given: $n(A) = 190, n(B) = 220, n(C) = 220$,$n

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$0.80$

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(A) $n(U) = 900$Let $A \equiv 	ext{Fever}$,$B \equiv 	ext{Cough}$,$C \equiv 	ext{Breathing problem}$.Given: $n(A) = 190, n(B) = 220, n(C) = 220$,$n(A \cup B) = 330, n(B \cup C) = 350, n(A \cup C) = 340, n(A \cap B \cap C) = 30$.Using $n(A \cup B) = n(A) + n(B) - n(A \cap B)$:$330 = 190 + 220 - n(A \cap B) \Rightarrow n(A \cap B) = 80$.Similarly,$350 = 220 + 220 - n(B \cap C) \Rightarrow n(B \cap C) = 90$.And $340 = 190 + 220 - n(A \cap C) \Rightarrow n(A \cap C) = 70$.Now,$n(A \cup B \cup C) = (n(A) + n(B) + n(C)) - (n(A \cap B) + n(B \cap C) + n(A \cap C)) + n(A \cap B \cap C)$$= (190 + 220 + 220) - (80 + 90 + 70) + 30 = 630 - 240 + 30 = 420$.Number of persons without any symptom $= n(U) - n(A \cup B \cup C) = 900 - 420 = 480$.Number of persons with exactly one symptom $= (n(A) + n(B) + n(C)) - 2(n(A \cap B) + n(B \cap C) + n(A \cap C)) + 3n(A \cap B \cap C)$$= (190 + 220 + 220) - 2(80 + 90 + 70) + 3(30) = 630 - 480 + 90 = 240$.Number of persons with at most one symptom $= 480 + 240 = 720$.Probability $= \frac{720}{900} = \frac{8}{10} = 0.80$.

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