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Two visitors are arriving at a house at the same time. Current population wide prevalence of infection is 0.01, i.e. 1%

The visitors are arriving from different destinations, and their infection status are independent of each other. No one in the house is infected.

Total probability of having someone infected in the house should be 0.01 + 0.01 = 0.02 given visitors are independent.

However, the probability of both visitors being infected is 0.01 x 0.01 = 0.0001 , again, given they're independent.

Should we use conditional probability here? If so, how? Both calculations apply to the real life situation at hand, so how do we apply fundamental probability calculations to real life here?

Update: In addition to Henry's answer, anybody arriving at this question via google may find the following blog post from scientific American also helpful. The authors are following the same approach Henry is suggesting.

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    $\begingroup$ "Total probability of having someone infected in the house should be 0.01 + 0.01 = 0.02 given visitors are independent" is not correct: independent events multiply probabilities. You might find it easier to find the probability nobody is infected $\endgroup$
    – Henry
    Jul 19, 2021 at 8:13
  • $\begingroup$ Thanks. That's helpful. Would you care to write an answer, maybe which prob. I should use for 'nobody is infected'? $\endgroup$ Jul 19, 2021 at 8:16

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"Total probability of having someone infected in the house should be $0.01 + 0.01 = 0.02$ given visitors are independent" is not correct: independent events multiply probabilities, while mutually exclusive events add probabilities.

You might find it easier to find the probability nobody is infected. That is $0.99 \times 0.99 = 0.9801$. So the probability someone is infected is $0.0199$.

You have already found the probability both are infected as $0.01 \times 0.01 = 0.0001$. The probability exactly one is infected is $0.01 \times 0.99 + 0.99\times 0.01=0.0198$. As you might hope, $0.0001+0.0198=0.0199$, so that give another approach to the same answer.

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  • $\begingroup$ Brilliant. Thanks. $\endgroup$ Jul 19, 2021 at 8:24

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