Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose a-priori chance of getting malaria is 10%. A positive blood test indicates a 80% chance of actually having the disease; but 5% of time healthy people also test positive. Suppose you test positive for malaria. What is the chance that u actually have the disease?

share|improve this question
add comment

1 Answer

up vote 0 down vote accepted

Let $M$ be the event the person has malaria, and let $P$ be the event the person tests positive. We want $\Pr(M|P)$. By the usual formula for conditional probability, we have $$\Pr(M|P)=\frac{\Pr(M\cap P)}{\Pr(P)}.$$ We want to find the two probabilities on the right.

First we deal with $\Pr(P)$. We can test positive in two ways: (i) the person has malaria and tests positive or (ii) the person does not have malaria but tests positive. These two events are disjoint. So to find $\Pr(P)$, we find the probabilities of (i) and of (ii), and add up.

We first find the probability of (i). The probability of having malaria is $0.10$. Given one has malaria, the probability of testing positive is $0.80$. So the probability of (i) is $(0.10)(0.80)$.

Similarly, the probability of (ii) is $(0.90)(0.05)$. Thus $$\Pr(M\cap P)=(0.10)(0.80)+(0.90)(0.05).$$

We have already computed $\Pr(M\cap B)$: it is just the probability of (i). The rest is straightforward computation.

Remark: We end up with $\Pr(M|P)=0.64$. So the malaria test is less impressive than a casual reading would indicate. If the incidence of malaria was $1\%$ rather than $10\%$, a substantial majority of positives would be false positives.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.