Question

In: Statistics and Probability

The prevalence of a disease D in the population is 10%. A new test is introduced...

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))

Solutions

Expert Solution

(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


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