# H1N1 question probability

A diagnostic test for H1N1 virus infection is 95 percent accurate, in that if a person is infected with the H1N1 virus, the test will detect it with a probability of 0.95, and if a person is not infected with the H1N1 virus, the test will give a negative result with a probability of 0.95. Suppose that only 0.5% of the population is infected with the H1N1 virus. One person is chosen at random from this population. The diagnostic test indicates that this person is infected. What is the probability that this person is actually not infected?

So far, all I have is: $$P(\text{positive test} \,|\, \text{truly infected}) = 0.95\\ P(\text{negative test}\, |\, \text{truly infected}) = 0.05\\ P(\text{H1N1 infected}) = 0.005 \implies P(\text{not H1N1 infected}) = 0.995?$$

Do I have it correct so far? I get stuck here... We are supposed to find $P(\text{positive test} \,| \,\text{not infected})$, correct?

Thank you for help in advance guys!

• You're correct so far, except that we're after P(not infected | positive test). Heard of Bayes' theorem? Commented Sep 28, 2017 at 6:49
• Yes, I understand the Bayes's theorem but I always get confused when it comes to conditional probability. Anyways, from what you hinted me I let $P(not infected) = P(H1N1 not infected)$ from the question above and $P(truly infected) = P(H1N1 infected)$. So I have: $P( not infected|positive test) = (P(positive test|not infected) P(not infected))/(P(positive test|truly infected) P(truly infected) + P(positive test|not infected) P(not infected))$ Commented Sep 28, 2017 at 7:09
• Just a suggestion, you could write $T, \bar{T},I, \bar{I}$ for respectively positivetest, negativetest, infected, notinfected. The formulas will be much easier to read, and look more like the Bayes theorem as it is written on wiki or in your reference book - which makes it easier to apply. Commented Sep 28, 2017 at 7:22

$$1 - \frac{0.5\% \times 95 \%}{0.5\% \times 95 \% + 99.5\% \times 5\%} = 1-\frac{95}{95+995} = 0.9128$$