Answer to Question #143756 in Statistics and Probability for qwerty

Question

1. In a certain region of the country, it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the probability of incorrectly diagnosing a person with cancer as having the disease is 0.06,

(a) what is the probability that an adult over 40 years of age is diagnosed as having cancer? (Give answer in decimal form in thousandths place. Ex. 0.031)
(b) What are the chances that a person has no cancer but is still diagnosed otherwise? (Give answer in percentage form. Please include percentage sign. Decimal must be in the thousandths place. Ex. 99.999%)

Answer 1

Let C denote the probability of selecting an adult over 40 years of age with cancer:

$P(C) = 0.05$ .

Hence C' represents the event that the person does not have a cancer:

$P(C') = 1 – P(C) = 0.95$ .

Let D denote the event that a person is diagnosed cancer. Then the probability of a doctor correctly diagnosing a person with cancer as having the disease:

$P(D|C) = 0.78$

and the probability of incorrectly diagnosing a person with cancer as having the disease:

$P(D|C') = 0.06$ .

a) The probability that an adult over 40 years of age is diagnosed as having cancer:

$P(D) = P(D|C)\cdot P(C) + P(D|C')\cdot P(C')$

$P(D)= 0.78\cdot0.05+0.06\cdot0.95=0.096$

b) The chances that a person has no cancer but is still diagnosed otherwise:

$P(C'|D)=\frac{P(D|C')\cdot P(C')}{P(D|C)\cdot P(C) + P(D|C')\cdot P(C')}$

$P(C'|D)=\frac{0.06\cdot0.95}{0.78\cdot0.05+0.06\cdot0.95}=\frac{0.057}{0.096}=0.59375=59.375\%$

Answer:

a) 0.096

b) 59.375%