n(U)=900 Let A≡ Fever, B≡ Cough C≡ Breathing problem ∴n(A)=190,n(B)=220,n(C)=220 n(A∪B)=330,n(B∪C)=350 n(A∪C)=340,n(A∩B∩C)=30 Now n(A∪B)=n(A)+n(B)−n(A∩B) ⇒330=190+220−n(A∩B) ⇒n(A∩B)=80 Similarly, 350=220+220−n(B∩C) ⇒n(B∩C)=90 and 340=190+220−n(A∩C) ⇒n(A∩C)=70 ∴n(A∪B∪C)=(190+220+220)−(80+90+70)+30
=660−240=420 ⇒ Number of person without any symptom =n(∪)−n(A∪B∪C) =900−420=480 Now, number of person suffering from exactly one symptom
=(n(A)+n(B)+n(C))−2(n(A∩B)+n(B∩C)+n(C∩A))+3n(A∩B∩C) =(190+220+220)−2(80+90+70)+3(30) =630−480+90=240 ∴ Number of person suffering from atmost one symotom =480+240=720 ⇒ Probability =