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.