Practice Problems

Problem 3:

I law of total probability and bayes rule

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Moderate

A test for coronaviruses has the following properties: Given that a patient has the coronavirus, the conditional probability of a positive test is .95. Given that a patient does not have the coronavirus, the conditional probability of a negative test is .95 You believe that in the population each patient has a .10 probability of having coronavirus. What is the probability that a person has coronavirus, given they present a positive result?

Solution:

Let $T$ be the event that a test comes up positive. Let $C$ be the event that an individual has coronavirus.

The problem tells us that

$P(C) = 0.1$
$P(T|C) = .95$
$P(T^C|C^C) = .95$

First, by the complement rule, $P(T|C^C) = 1 - P(T^C|C^C) = 1 - .95 = .05$. Also by the complement rule, $P(C^C) = 1-P(C) = 1 - 0.1 = 0.9$

To apply Bayes' Theorem, we first need the probability of a positive test. We notice that ${C, C^C}$ is a partition of the sample space. Writing the law of total probability,

$$P(T) = P(C)P(T|C) + P(C^C)P(T|C^C) = (0.1)(0.95) + (0.9)(0.05) = 0.14$$

Finally, we write down Bayes' Theorem:

$$P(C|T) = \frac{P(T|C)P(C)}{P(T)} = \frac{(0.95)(0.1)}{0.14} = 0.68$$.

Note: As is often the case in problems like this, the test appears very accurate, but observing a positive result doesn't lead to a very high posterior probability of having the disease.