Question

In a certain region of the country it is known from

pastexperience that theprobability of selecting an adult over 40
yearsof age with cancer is 0.05. If the probability of a
doctorcorrectly diagnosing a person with cancer as having the
disease is0.78 and the probability of incorrectly diagnosing a
person withoutcancer as having the disease is .06, what is the
probability that aperson is diagnosed as having cancer?

Answer 1

There is a 9.6% probability that a person is diagnosed as having cancer.

Explanation:

In this problem, we have these following probabilities:

A 5% probability that an adult over 40 has cancer.

This also means that:

There is a 95% probability that an adult over 40 does not have cancer. (Since either the adult has cancer or does not have cancer, and the sum of the probabilities is 100%).

A 78% probability of a person that has cancer being diagnosed,

A 6% probability of a person that does not have cancer being diagnosed.

What is the probability that a person is diagnosed as having cancer?

`P = P_(1) + P_(2)`

`P_(1)` is the probability of those who have cancer being diagnosed. So it is 78% of 5%. So

`P_(1) = 0.05*0.78 = 0.039`

`P_(2)` is the probability of those who do not have cancer being diagnosed. So it is 6% of 95%. So

`P_(1) = 0.06*0.95 = 0.057`

So

`P = P_(1) + P_(2) = 0.039 + 0.057 = 0.096`

There is a 9.6% probability that a person is diagnosed as having cancer.