Question

. The probability that a patient catches flu is 3/8, the probability that he has a headache
is 5/15, and the probability that he has asthma is 4/9. If we know the fact that the
probability that the patient will have at least one of the illnesses is 3/5, then what is
the probability that he will have flu, headache, and asthma?

Answer 1

Answer:

$\boxed {\frac{343}{360}}$

Step-by-step explanation:

Probability Rule :

$\boxed {P(A\cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B \cap C)}$

Solving :

⇒ 3/5 = 3/8 + 5/15 + 4/9 - P (A ∩ B ∩ C)

⇒ P (A ∩ B ∩ C) = 3/8 + 1/3 + 4/9 - 1/5

⇒ P (A ∩ B ∩ C) = 135/360 + 120/360 + 160/360 - 72/360

⇒ P (A ∩ B ∩ C) = (135 + 120 + 160 - 72) / 360

⇒ P (A ∩ B ∩ C) = 343/360