A young man visits a hospital for medical check-up. The probability that he has lung problem is 0.45, heart problem is 0.29 and either lungs or heart problem is 0.47. What is the probability that he has both types of problems: lungs as well as heart? Out of 1000 people, how many are expected to have both types of problem? What should we do to maintain good health and keep away from hospital? Describe briefly. Share with your friends Share 12 Arjit Singh answered this P(L)= 0.45P(H) =0.29P(L∩H)=0.47Now we know the relationship:P(L∪H) = P(L)+P(H)-P(L∩H)=0.27If in one person, the probability of having both diseases is 0.27then in 1000 persons, 0.27*1000=270 are expected to have both the diseases.Though not a mathematics question, to maintain good health following points can be kept in mind:1. Proper diet with balance of all micronutrients like vitamins, proteins, etc2. Proper sleep cycle3. Physical exercise and yoga and meditation4. Clean surroundings 15 View Full Answer Unique Pearl answered this Let the probability of having a lung problem be P ( L )Let the probability of having a heart problem be P ( H )As per the given question , P ( L ) = 0.45 and P ( H ) = 0.29 Either lungs or heart problem => P ( H U L ) = 0.47Probabilty of having both types of problems => P ( H ^L ) = ?P ( H U L ) = P ( H ) + P ( L ) - P ( H ^ L )0.47 = 0.29 + 0.45 - xx = 0.27 ie P ( H ^ L ) = 0.27 And u can add some biology precautions too :P Lazy to type :P 3
