Question

A certain medical disease occurs in 5% of the population. A simple screening procedure is available and in 3 out of 10 cases where the patient has the disease it provides a positive result. If the patient does not have the disease there is still a 0.05 chance that the test will give a positive result. Draw a tree diagram to represent this.
a) Find the probability that a randomly selected individual
does not have the disease but gives a positive result in the screening test
gives a positive result on the test.
b) Ben has taken the test and his result is positive. Find the probability that he has the disease

Answer 1

Answer on Question #59251 – Math – Statistics and Probability

Question

A certain medical disease occurs in 5% of the population. A simple screening procedure is available and in 3 out of 10 cases where the patient has the disease it provides a positive result. If the patient does not have the disease there is still a 0.05 chance that the test will give a positive result. Draw a tree diagram to represent this.

a) Find the probability that a randomly selected individual does not have the disease but gives a positive result in the screening test gives a positive result on the test.

b) Ben has taken the test and his result is positive. Find the probability that he has the disease.

Solution

A tree diagram is given below.

Introduce $H_{1} =$ "disease occurs", $H_{2} =$ "disease does not occur", $D =$ "positive result",

$$P(H_{1}) = 0.05, \quad P(H_{2}) = 0.95, \quad P(D|H_{1}) = 0.3, \quad P(D|H_{2}) = 0.05.$$

a) By the rule of multiplication, the probability that a randomly selected individual does not have the disease but gives a positive result in the screening test is

$$P_{1} = P(H_{2} \cap D) = P(D|H_{2}) \cdot P(H_{2}) = 0.05 \cdot 0.95 = 0.0475 \quad \text{or} \ 4.75\%.$$

b) We shall apply Bayes' Theorem below.

Given his result is positive the probability that he has the disease is

$$P(H_{1}|D) = \frac{P(D|H_{1}) \cdot P(H_{1})}{P(D)} = \frac{P(D|H_{1}) \cdot P(H_{1})}{P(D|H_{1}) \cdot P(H_{1}) + P(D|H_{2}) \cdot P(H_{2})} = \frac{0.3 \cdot 0.05}{0.3 \cdot 0.05 + 0.05 \cdot 0.95} = 0.24 \quad \text{or} \ 24\%.$$

Answer: a) 0.0475; b) 0.24.

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