Question

A new, simple test has been developed to detect a particular type of cancer. The test must be evaluated before it is used. A medical researcher selects a random sample of 1,000 adults and finds (by other means) that 2% have this type of cancer. Each of the 1,000 adults is given the test, and it is found that the test indicates cancer in 98% of those who have it and in 1% of those who do not. Based on these results, what is the probability of a randomly chosen person not having cancer given that the test indicates cancer

Answer 1

Answer:

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer

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: Test indicates cancer.

Event B: Person does not have cancer.

Probability of a test indicating cancer.

98% of 2%(those who have).

1% of 100 - 2 = 98%(those who do not have). So

$P(A) = 0.98*0.02 + 0.01*0.98 = 0.0294$

Probability of a test indicating cancer and person not having.

1% of 98%. So

$P(A \cap B) = 0.01*0.98 = 0.0098$

What is the probability of a randomly chosen person not having cancer given that the test indicates cancer?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0098}{0.0294} = 0.3333$

0.3333 = 33.33% probability of a randomly chosen person not having cancer given that the test indicates cancer