Question

Assume a test for a disease has a probability 0.05 of incorrectly identifying an individual as infected (False Positive), and a probability 0.01 of incorrectly identifying an individual as uninfected (False Negative). Assume furthermore that the probability of an individual being infected is 0.000001 (1 in 1 million). If a person tests positive for this disease, what is the probability of actually having the disease

Answer 1

Answer:

0.00002 = 0.002% probability of actually having the disease

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Positive test

Event B: Having the disease

Probability of having a positive test:

0.05 of 1 - 0.000001(false positive)

0.99 of 0.000001 positive. So

$P(A) = 0.05*(1 - 0.000001) + 0.99*0.000001 = 0.05000094$

Probability of a positive test and having the disease:

0.99 of 0.000001. So

$P(A \cap B) = 0.99*0.000001 = 9.9 \times 10^{-7}$

What is the probability of actually having the disease

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{9.9 \times 10^{-7}}{0.05000094} = 0.00002$

0.00002 = 0.002% probability of actually having the disease