# Show that the probability that a person actually has this life threatening disease, if they test positive to the blood test is actually $0.1248905$.

I am looking for help with this seemingly simple probability problem:

Suppose $20,000$ Australians have a rare life-threatening disease (that is, about $0.1\%$ of the population). The following table shows approximate probabilities of positive and negative outcomes from the a blood test of detection for this disease.

                   Blood Test     +      -
Antibodies Present   0.998  0.002
Antibodies Absent    0.007  0.993


a) Show that the probability that a person actually has this life threatening disease (i.e., antibodies present), if they test positive to the blood test, is actually $0.1248905$ ($12.48905\%$). In your answer, define appropriate events of interest and conditional probabilities in terms of these events.

Attempt:

My original thought was to compute this as a simple conditional probability. So, if we let $$A=P(\text {A person has the disease})$$ and $$B=P(\text {A person tested positive on the blood test})$$ then we want to find $P(A|B)$ (i.e. given a person has tested positive for the antibodies what is the probability they actually have the disease). However, when I do the calculation I see that \begin{align*} P(A|B)&=\dfrac{P(A\cap B)}{P(A)}\\ &=\dfrac{P(A)\cdot P(B)}{P(A)}\\ &= P(B)\\ &=0.998. \end{align*} Clearly this is wrong as the question asked to show the probability is $0.1248905$.

I therefore must conclude that my logic is faulty. I am perhaps thinking that I assumed the two events are independent when they are actually not independent, but I still seem to get nowhere.

Please can someone help to point me in the right direction with this question.

• en.wikipedia.org/wiki/Bayes'_theorem#Drug_testing – amd Mar 22 '16 at 5:47
• The correct formula is $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$, so that if $A$ and $B$ are independent then $P(A\mid B) = P(A)$. (But they are not independent, so you can't even say that.) Even without considering the number that was prompted you should notice that "$P(A\mid B) = P(B)$" is not right. – David K Mar 22 '16 at 12:04
• I find it amusing that a question that starts by saying that "about $0.01\%$ of the population" has the disease ends by asking you to prove that a certain person has it is $12.48905\%$. That's an awfully precise outcome for such imprecise input. Of course you should still answer the question that was asked, I'm just pointing out that it doesn't quite work that way in the real world. – David K Mar 22 '16 at 12:10

Let $A^\pm$ be the event that the person does or does not have the disease, and let $B^\pm$ be the event that they tested $+$ or $-$. Hence, we have \begin{align*} P(A^+|B^+) &= \frac{P(A^+\cap B^+)}{P(B^+)}\tag 1\\ &=\frac{P(B^+|A^+)P(A^+)}{P(A^+\cap B^+)+P(A^-\cap B^+)}\tag 2\\ &=\frac{P(B^+|A^+)P(A^+)}{P(B^+|A^+)P(A^+)+P(B^+|A^-)P(A^-)} \end{align*} where in $(1)$ I used Bayes' rule and in $(2)$ I used the product rule on the numerator, and the law of total probability on the denominator.