00:01
In this exercise, it's given that the probability that somebody has a certain disease is 0 .04.
00:07
So i've denoted the event in which somebody has a disease as d.
00:12
So it's given the probability of d is 0 .04.
00:16
There is a test for this disease, and we are told that the probability that the test comes out positive, given that the disease is present, is 0 .88.
00:25
So this probability is 0 .88.
00:27
And the probability of a positive test if the person does not have the disease is 0 .02.
00:35
For part a, we were asked, if the test comes back positive, what is the probability that the disease is actually present? so this is a conditional probability, probability that the disease is present, given that the test came back positive.
00:54
Now we can use bays ' theorem to help solve this.
01:01
Bases theorem basically says probability of b given a, is equal to the probability of a given b times the probability of b divided by the probability of a.
01:16
So for our scenario, this is equal to the probability of testing positive, given that the disease is present, times the probability of d divided by the probability of testing positive.
01:30
Now for the probabilities in the numerator here, we have both of these given in the question, but we need to solve the probability of testing positive.
01:38
So let's do that as an intermediate calculation.
01:42
To do this, we can use the law of total probability.
01:45
So there's two ways to test positive here...