Question

2. We want to market a new test for disease detection. 99 of 100 people who have the specific disease tested positive for that disease. 2 of 100 people who do not have the disease tested positive. If we assume 10% of the total population currently has this disease, find the probability a randomly selected person has the disease if he tests positive for that disease.

probabilities I know from given problem:

.99 have disease AND Test + therefore...
.01 have disease AND Test -

.02 do not have disease AND Test + therefore...
.98 do not have disease AND Test -

.10 of TOTAL population HAVE Disease therefore...
.90 of TOTAL population DO NOT HAVE Disease.

what I thought I would have to do to get what is being asked is P(have disease | tests +) = P(Have disease AND Test +) / P(test +) = .99 / p(test +). however I thought p(test +) was the .10 that has the disease but .99 / .10 is 9.9 which is greater than 1 and a probability cannot be greater than 1 so I am confused what the pictured question is asking.

thanks for your help!

Answer 1

Solution:-

$P(\frac{Disease}{Positive}) = 0.8462$

P(Disease) = 0.10

P(Not disease) = 1 - 0.10 = 0.90

$P(\frac{Positive}{Disease}) = 0.99$

$P(\frac{Positive}{Not.disease}) = 0.02$

$P(\frac{Disease}{Positive}) =\frac{P(Disease)P(\frac{Positive}{Disease})}{P(Disease)P(\frac{Positive}{Disease})+P(Not .disease)P(\frac{Positive}{Not .disease})}$

$P(\frac{Disease}{Positive}) = \frac{0.10\ast 0.99}{0.10\ast 0.99+0.90\ast 0.02}$

$P(\frac{Disease}{Positive}) = 0.8462$