Question

You are working in a primary care office. Flu season is starting. For the sake of public health, it is critical to diagnose people with the flu. The prior probability that someone who is walking through your door has the flu is 0.1. If someone has the flu, their probability of having a runny nose is 0.99. However, if someone doesn’t have the flu, e.g. just has the cold (they will (think that they) have something, otherwise they wouldn’t seek out your office), their probability of having a runny nose is 0.9. Someone comes in and has a runny nose. What is the probability that this person has the flu? A. 0.5 B. 1 C. 0.1 D. 0 E. 0.11

Answer 1

Answer:

E. 0.11

Step-by-step explanation:

We have these following probabilities:

A 10% probability that a person has the flu.

A 90% probability that a person does not have the flu, just a cold.

If a person has the flu, a 99% probability of having a runny nose.

If a person just has a cold, a 90% probability of having a runny nose.

This can be formulated as the following problem:

What is the probability of B happening, knowing that A has happened.

It can be calculated by the following formula

$P = \frac{P(B).P(A/B)}{P(A)}$

Where P(B) is the probability of B happening, P(A/B) is the probability of A happening knowing that B happened and P(A) is the probability of A happening.

In this problem, we have that:

What is the probability that a person has the flu, given that she has a runny nose?

P(B) is the probability that a person has the flu. So P(B) = 0.1.

P(A/B) is the probability that a person has a runny nose, given that she has the flu. So P(A/B) = 0.99.

P(A) is the probability that a person has a runny nose. It is 0.99 of 0.1 and 0.90 of 0.90. So

$P(A) = 0.99*0.1 + 0.9*0.9 = 0.909$

What is the probability that this person has the flu?

$P = \frac{P(B).P(A/B)}{P(A)} = \frac{0.1*0.99}{0.909} = 0.1089 = 0.11$

The correct answer is:

E. 0.11