# Need help with a probability problem

Hi All,

I need a quick help with this problem i have been working with. I know the answer (from book). My question is how do I reach that answer. Now im not an expert in probability but I tried every way i can think of but to no avail.

The Answer in the Book: 0.0960

Question:
In a certain region of the country it is known from past experience that the probability of selecting an adult over 40 years of age with cancer is 0.05. If the probability of a doctor correctly diagnosing a person with cancer as having the disease is 0.78 and the
probability of incorrectly diagnosing a person without cancer as having the disease is 0.06, what is the probability that an adult over 40 years of age is diagnosed as having cancer?

Any help will be much appreciated.

Hi, blondii. I'll start with a disclaimer stating that I know next to nothing about probability mathematics, and very little about mathematics in general, but here's the solution that makes sense to me:

In this area, it is assumed that 5% of people have cancer and 95% do not. Of the 5% who do, a doctor will make a correct diagnosis 78% of the time.

.78 * .05 = .039

So there is a 3.9% chance that a person will a) have cancer, and b) receive an accurate diagnosis.

Of the remaining 95%, 6% of them will receive a false positive diagnosis.

.95 * .06 = .057

The combined probabilities of the two groups is 9.6%.

Hope that helps,
Pete

Hi SneakyPete, thanks for your solution. Seems to make sense. Much appreciated.