Anew test to detect tb has been designed. it is estimated that 89% of people taking this test have the disease. the test detects the disease in 97% of those who have the disease. the test does not detect the disease in 99% of those who do not have the disease. if a person taking the test is chosen at random, what is the probability of the test indicating that the person does not have the disease?

0.1356

Explanation:

This is a probability exercise. Let's define some probability concepts.

Given two events A and B :

(A∩B) = (A,B)

(A,B) is the intersection event where A and B occur both at the same time.

We define as the conditional probability '' The probability of the event A given that we know that the event B occurred'' as :

Where

Now, if A is an event and is its complement ⇒

Finally we define the probability of the union between two events A and B :

P(A∪B) = P(A) + P(B) - P(A,B)

If the events A and B are independent between them ⇒ P(A,B) = 0 ⇒

P(A∪B) = P(A) + P(B)

Let's define the following events for this exercise :

D : ''People taking this test that have the disease''

P : ''The test is positive''

And

We are looking the probability of

P(P') = P [(P'∩D) ∪ (P'∩D')]

Given that this events are independent between them :

(I)

Let's write the conditionals for this problem :

⇒

(II)

And the another conditional :

⇒

(III)

Replacing (II) and (III) in (I) :

We find that the probability of the test indicating that the person does not have the disease ( P(P') ) is 0.1356

answer:

noting will happen take it easy ..

explanation:

don't be too much serious or worried nothing will happen to you .

drink a glass of water and maybe . just keep on roaming for a while so that the water is absorbed easily and don't forget to drink plenty of water .okk don't worry you will be fine

the answer is d

hope this and you do good