Answer:
a) 65%
b) 28.57%
Step-by-step explanation:
Suppose that 35% of all patients admitted to a hospital's
intensive care unit have high blood pressure, 42% have some sort of infection, and 12% have both problems. Probabilities estimated from the data in ICU Admissions. (10POINTS)
(a) Find the probability that a randomly chosen patient in this ICU has either high blood pressure or an infection.
(b) Find the probability that a patient admitted to the ICU has high blood pressure given that an infection is present.
Solution:
Let H represent the people with high blood pressure and I represent the people with infection.
P(H) = 35% = 0.35, P(I) = 42% = 0.42, P(H ∩ I) = 12% = 0.12
Those with only high blood pressure = P(H ∩ I') = P(H) - P(H ∩ I) = 0.35 - 0.12 =0.23
Those with only infection = P(H' ∩ I) = P(I) - P(H ∩ I) = 0.42 - 0.12 =0.30
a) probability that a patient has either high blood pressure or an infection = P(H ∪ I)
P(H ∪ I) = P(H ∩ I') + P(H' ∩ I) + P(H ∩ I) = 0.23 + 0.3 + 0.12 = 0.65 = 65%
b) probability that a patient has high blood pressure given that an infection is present = P(H / I)
P(H / I) = P(H ∩ I) / P(I) = 0.12 / 0.42 = 0.2857 = 28.57%
since it is exactly in the halfway between 