# Thread: Basic Probability help

1. ## Basic Probability help

It is known that the probability of selecting an adult over 40 years of age with cancer is 0.05. 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 a person diagnosed as having cancer actually has the disease?

I'm a bit of a loss as far as setting up this problem goes. The probability of a person diagnosed as having cancer is Pr(A) = 0.78+0.06 = 0.84, right? And we know that the probability of a person having cancer is Pr(B) = 0.05.

So, I would guess that I'm looking for Pr(A|B) = (0.84*0.05)/0.05 = 0.84

That just doesn't seem right. What am I doing wrong?

2. I think you need to use the law of total probability.
Let $A$ be the probability that the person has cancer, so $\bar{A}$ is the probability that the person doesn't have cancer.
And $D$ the probability that the person gets diagnosed, then;

$P(D)=P(A)P(D|A)+P(\bar{A})P(D|\bar{A})=.05\cdot .78 + .95\cdot .06=.096$

hmm, actually, i dont know where i was going with this :P let me think about it a little more...

3. I think I needed to use Bayes' theorem. With it, I got an answer around 0.4, which seems kind of low, but I don't know. I'll post all of the math in a bit.

A = diagnosed positive
B = has cancer

Pr(B) = 0.05
Pr(A|B) = 0.78
Pr(A|notB) = 0.06

Pr(A) = [Pr(A|B)*Pr(B)] + [Pr(A|notB)*Pr(notB)] = (0.78*0.05)+(0.06*0.95) = 0.096

Pr(B|A) = [Pr(A|B)*Pr(B)]/[(Pr(A|B)*Pr(B)) + Pr(A|notB)*Pr(notB)] =
(0.78*0.05)/[0.78*0.05)+(0.06*0.95)] = 0.039/0.096 = 0.40625

Can anyone confirm that I did this right? Sorry about the syntax. I'm not familiar with the proper formatting when using the MATH tags. I'll need to work on learning that for future posts.

4. Yes, you will need to use Bayes theorem;
I wasn't reading the question clearly, you want the pr someone has cancer, given they were diagnosed with cancer.

$P(A|D)=\frac{P(A)\cdot P(D|A)}{P(D)}=\frac{.05\cdot .78}{.096}=.40625$

Which i am guessing looks familiar?

5. Awesome. Thanks a lot