# Math Help - Joint Probability

1. ## Joint Probability

From past experience, 10 % of the people over 30 years have a certain type of disease. The probability of a doctor correctly diagnosing a person with the disease is 78% and incorrectly not having the disease 8%.

a. what is the probability that a person is diagonsed as having the disease?

b. what is the probability that a person diagonesd as having the disease actually have the disease?

2. Originally Posted by weakmath
From past experience, 10 % of the people over 30 years have a certain type of disease. The probability of a doctor correctly diagnosing a person with the disease is 78% and incorrectly not having the disease 8%.

a. what is the probability that a person is diagonsed as having the disease?

b. what is the probability that a person diagonesd as having the disease actually have the disease?
Let D be the event "have the disease".
Let A be the event "diagnosed with the disease".

Given data:

"10 % of the people over 30 years have a certain type of disease": Pr(D) = 0.1 => Pr(D') = 0.9

"The probability of a doctor correctly diagnosing a person with the disease is 78%": Pr(A | D) = 0.78 => Pr(A' | D) = 1 - 0.78 = 0.22

"The probability of a doctor ... incorrectly not having the disease 8%": Pr(A | D') = 0.08 => Pr(A' | D') = 0.92 (well, this is the only interpretation of the wording that makes any sense to me, anyway)

a. Pr(A) = Pr(A | D) Pr(D) + Pr(A | D') Pr(D') = ....

b. You need Pr( D | A).

Note that $\Pr(D | A) \cdot \Pr(A) = \Pr(D \cap A)$ and $\Pr(A | D) \cdot \Pr(D) = \Pr(A \cap D) = \Pr(D \cap A)$.

Therefore $\Pr(D | A) \cdot \Pr(A) = \Pr(A | D) \cdot \Pr(D) \Rightarrow \Pr(D | A) = \frac{\Pr(A | D) \cdot \Pr(D) }{\Pr(A) }$.

This is just another way of writing the usual formula $\Pr(D | A) = \frac{\Pr(D \cap A)}{\Pr(A)}$.

Now substitute the values of Pr( A | D), Pr(D) and Pr(A).