A young man visits a hospital for medical checkup. The probability tha

1.	A young man visits a hospital for medical checkup. The probability that he has lungs problem is 0.55, heart problem is 0.29 and either lungs or heart problem is 0.57. What is the probability that he has (i) both type of problems: lungs as well as heart? (ii) lungs problem but not heart problem? Out of 100 person, how many are expected to have both type of problems?
Answer» Let ‘L’ and ‘H’ be the events that the young man has lungs problem and heart problem respectively. ∴ P(L) = 0.55, P(H) = 0.29, P(∪ H) 0 = 0.57 P(L ∩ H) = ? (i) We know that, P(L ∪ H) = P(L) + P(H) – P(L ∩ H) ⇒ P(L ∩ H) = P(L) + P(H) – P(L ∩ H) = 0.55 + 0.29 – 0.57 = 0.27 ∴ The probability that he has both the problems is 0.27. (ii) p(L ∩ H’) = p(L) – p(L ∩ H) = 0.55 – 0.27 = .28 ∴ The probability that he has lungs problem but not heart problem is 0.28 ∴ Out of 1000 persons, the number of persons having both the problems = 0.27 × 1000 = 270.32

A young man visits a hospital for medical checkup. The probability that he has lungs problem is 0.55, heart problem is 0.29 and either lungs or heart problem is 0.57. What is the probability that he has (i) both type of problems: lungs as well as heart? (ii) lungs problem but not heart problem? Out of 100 person, how many are expected to have both type of problems?

Answer»

Let ‘L’ and ‘H’ be the events that the young man has lungs problem and heart problem respectively.

∴ P(L) = 0.55, P(H) = 0.29, P(∪ H) 0 = 0.57

P(L ∩ H) = ?

(i) We know that,

P(L ∪ H) = P(L) + P(H) – P(L ∩ H)

⇒ P(L ∩ H) = P(L) + P(H) – P(L ∩ H)

= 0.55 + 0.29 – 0.57

= 0.27

∴ The probability that he has both the problems is 0.27.

(ii) p(L ∩ H’) = p(L) – p(L ∩ H)

= 0.55 – 0.27

= .28

∴ The probability that he has lungs problem but not heart problem is 0.28

∴ Out of 1000 persons, the number of persons having both the problems

= 0.27 × 1000 = 270.32