Question

An experimental drug has been shown to be 70% effective in eliminating symptoms of allergies in animal studies. A small human study involving 9 participants is conducted. What is the probability that the drug is effective on at least 5 of the participants?

Individuals exhibiting a certain set of symptoms are screened for a viral infection. Suppose that: (i) the screening test results in a positive diagnosis for 75% of individuals who really do have the infection; (ii) the screening test results in a negative diagnosis for 75% of individuals who really do not have the infection; and (iii) 10% of individuals exhibiting the set of symptoms have the infection. Let + denote the event that a randomly selected individual exhibiting the set of symptoms is diagnosed positively, and let D denote the event that this individual has the disease.

What is the Pr(+|D)?

What is the Pr(+| )?

What is the Pr(D)?

What is the Pr(D|+)?

Answer 1

drug is effective on at least 5 of the participants = P(X=5) + P(X=6) + P(X=7) + P(X=8) +P(X=9)

= 9C5 * 0.7⁵ * 0.3⁴ + 9C6 * 0.7⁶ * 0.3³ + 9C7 * 0.7⁷ * 0.3² + 9C8 * 0.7⁸ * 0.3¹ + 9C9 * 0.7⁹ * 0.3⁰

=

X	9CX	0.7X	0.3(9-X)	9CX * 0.7X * 0.3(9-X)
5	126	0.16807	0.0081	0.172
6	84	0.117649	0.027	0.267
7	36	0.082354	0.09	0.267
8	9	0.057648	0.3	0.156
9	1	0.040354	1	0.040

total = 0.901

let probability of individual having symptoms of infection is = x

then probability of individual not having symptoms of infection is = 1 - x

+ denote the event that a randomly selected individual exhibiting the set of symptoms is diagnosed positively

P(+) = 0.75 * P(D) + 0.25 * (1-x) = 0.5 * x + 0.25

D denote the event that this individual has the disease = P(D)

probability of event that the individual does not have disease but showing symptom = 1 - P(D)

P(D) = 0.1

P(+) = has disease and diagnosed positibely + does not have disease but diagnosed positively = diagnosed positively irrespective of having the disease or not

= 0.1 * 0.75 + 0.9 * (1-0.75) = 0.075+0.225 = 0.3

P(+|D) = has disease and diagnosed properly / has disease = 0.1*0.75 / 0.1 = 0.75

P(D|+) = has disease and diagnosed properly / diagnosed positively = 1*0.75 / 0.3 = 0.25