# Probability of actually being sick after 3 positive tests? (Bayes problem?)

I was given the following problem:

The probability of being sick is $$0.05$$. Probability of a test detecting sickness when you're truly sick is $$0.8$$. Probability of detecting sickness when you're not sick is $$0.01$$. If you take 3 consecutive tests, and they all are positive, what is the actual probability of you being sick?

I think this is a Bayes Rule problem. Let $$S$$ denote that you're truly sick and $$S^c$$ denote that you're not sick. Let $$+$$ and $$-$$ denote positive and negative test, respectively. I think first I need to find $$P(S|+)$$, probability that you're sick given one positive test.

We know: \begin{align} P(+|S) = 0.8 \\ P(-|S) = 0.2 \\ P(+|S^c) = 0.01 \\ P(-|S^c) = 0.99 \\ \end{align}

According to Bayes rule, we have $$P(S|+) = \frac{P(+|S)P(S)}{P(+)}$$

Then I think I can find the marginals using: \begin{align} P(+) = P(+|S)P(S) + P(+|S^c)P(S^c) \\ P(S) = P(S|+)P(+) + P(S|-)P(-) \\ \end{align}

Assuming this is the correct approach, and I find $$P(S|+)$$, how does this generalize to the case where there are 3 consecutive positive tests, i.e., how do I find $$P(S|+,+,+)$$?

• You need to make some assumption regarding the independence of the test results. If, say, the test reports a false negative, is it more likely to report a false negative on the second try?
– lulu
Commented Aug 15, 2020 at 21:02
• @lulu I think we assume independency between the trials. Commented Aug 15, 2020 at 21:06
• So, in that case, what's the probability of a sick person getting three positive results? What's the probability of a healthy person getting three positive results?
– lulu
Commented Aug 15, 2020 at 21:12
• @lulu These are things I need to solve right, or are you asking for clarification? Commented Aug 15, 2020 at 21:14
• I'm saying that independence gives you the answer to those. Try an easier one: what's the probability that a sick person gets two positive results?
– lulu
Commented Aug 15, 2020 at 21:41

$$P(S|+) = \frac{P(+|S)P(S)}{P(+)}\\ P(+) = P(+|S)P(S) + P(+|S^c)P(S^c)$$
And you give probabilities at the start that enable you to find all the necessary probabilities, including $$P(S) = 0.05$$ [and so $$P(S^c) = 0.95]$$ in your first sentence. So you can get $$P(S|+),$$ which you say you need.
Why does your last displayed line try to compute $$P(S)?$$