# Probability question on conditional prob

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?

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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.

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