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 $P(S|+,+,+)$?

  • 2
    $\begingroup$ 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? $\endgroup$
    – lulu
    Aug 15, 2020 at 21:02
  • $\begingroup$ @lulu I think we assume independency between the trials. $\endgroup$
    – roulette01
    Aug 15, 2020 at 21:06
  • 1
    $\begingroup$ 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? $\endgroup$
    – lulu
    Aug 15, 2020 at 21:12
  • $\begingroup$ @lulu These are things I need to solve right, or are you asking for clarification? $\endgroup$
    – roulette01
    Aug 15, 2020 at 21:14
  • 1
    $\begingroup$ 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? $\endgroup$
    – lulu
    Aug 15, 2020 at 21:41

1 Answer 1


Comment: You say correctly:

$$ P(S|+) = \frac{P(+|S)P(S)}{P(+)}\\ P(+) = P(+|S)P(S) + P(+|S^c)P(S^c) $$

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 $P(S)?$

I agree with @lulu that you will have to assume that tests are independent to finish this.


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