Are the following equations equivalent why or why not Branilist for people who explain
1 answer:
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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
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
What is the probability that this person has the flu?
The correct answer is:
E. 0.11