Answer:
The percent of the people who tested positive actually have the disease is 38.64%.
Step-by-step explanation:
Denote the events as follows:
X = a person has the disease
P = the test result is positive
N = the test result is negative
Given:
[tex]P(X)=0.01\\P(P|X^{c})=0.15\\P(N|X)=0.10[/tex]
Compute the value of P (P|X) as follows:
[tex]P(P|X)=1-P(P|X^{c})=1-0.15=0.85[/tex]
Compute the probability of a positive test result as follows:
[tex]P(P)=P(P|X)P(X)+P(P|X^{c})P(X^{c})\\=(0.85\times0.10)+(0.15\times0.90)\\=0.22[/tex]
Compute the probability of a person having the disease given that he/she was tested positive as follows:
[tex]P(X|P)=\frac{P(P|X)P(X)}{P(P)}=\frac{0.85\times0.10}{0.22} =0.3864[/tex]
The percentage of people having the disease given that he/she was tested positive is, 0.3864 × 100 = 38.64%.
