00:01
Hi, i'm david and i'm here to help you and change the question.
00:03
Now let me bring up your question here.
00:05
In the question here, let me summarize the question by the 3 diagram.
00:09
We have the probability that the person has a certain dz equal to the subon 03.
00:13
So we will have dz denoted by the d and not having the dz will be the d prime.
00:19
And then we have here will be the z1 03, not having the dz will be 1 minus 03 equal to the zvon 97.
00:27
And then when they take the task, the result can be either the positive or the negative, the result will be either positive on the negative.
00:37
And we are told that if the disease actually present the probability to the medical test give the positive result equal to the 0188, then the common negative will be the 1, 2.
00:51
And if the disease not actually present the probability on the positive test results equal to the 0101, the common common will be the subon negative negative and that will be the summary of the question.
01:03
Now if the test shows the positive results, so probability given that would be the positive result, once you find the probability that the disease is actually present, it will be the probability of the d.
01:16
To find this probability, each and equal to by conditional probability equal to probability the probability and positive result...