In: Statistics and Probability
The prevalence of a disease D in the population is 10%. A new
test is introduced by a laboratory with the following
characteristics. The laboratory notes that if someone has the
disease, his or her test will be positive 90% of the time (P(pos|D)
= 0.90). However, a completely healthy individual will test
positive with probability 0.06 (P (pos|Dc) = .06).
(a) If we randomly select 5 people from this population, what is
the probability that all of them will be healthy?
(b) What is the probability that not more than 2 will be sick?
(c) Compute all the joint probabilities between the test and disease status[ P(D and pos), P(D and Neg), P(Dc and pos), and P(Dc and Neg) ]
(d) If a test is positive, what is the probability that the
patient is actually sick (P(D|pos))
(a)
Probability that a randomly selected person is healthy = P(Dc) = 1 - P(D) = 1 - 0.1 = 0.9
Probability that all 5 of them will be healthy = P(Dc)5 = 0.95 = 0.59049
(b)
Let X be the number of sick people in selected 5 persons. Then X ~ Binomial(n = 5, p = 0.1)
Probability that not more than 2 will be sick = P(X 2)
= P(X = 0) + P(X = 1) + P(X = 2)
= 5C0 * 0.10 * 0.95 + 5C1 * 0.11 * 0.94 + 5C2 * 0.12 * 0.93
= 0.59049 + 0.32805 + 0.07290
= 0.99144
(c)
P(D and pos) = P(pos|D) P(D) = 0.9 * 0.1 = 0.09
P(D and Neg) = P(Neg | D) P(D) = (1 - P(pos|D) ) P(D) = (1 - 0.9) * 0.1 = 0.01
P(Dc and pos) = P (pos|Dc) P(DC) = 0.06 * 0.9 = 0.054
P(Dc and Neg) = P(Neg | Dc) P(DC) = (1 - P (pos|Dc)) * P(DC)
= (1 - 0.06) * 0.9
= 0.846
(d)
By Law of total probability,
P(Pos) = P(pos|D) P(D) + P(pos | DC) P(DC)
= 0.90 * 0.1 + 0.06 * 0.9
= 0.144
P(D | pos) = P(pos|D) P(D) / P(pos) (By Bayes theorem)
= 0.90 * 0.1 / 0.144
= 0.625