Refer to Example 2.26.

a. If a man tests negative, what is the probability that he actually has the disease?

b. For many medical tests, it is standard procedure to repeat the test when a positive signal is given. If repeated tests are independent, what is the probability that a man will test positive on two successive tests if he has the disease?

c. Assuming repeated tests are independent, what is the probability that a man tests positive on two successive tests if he does not have the disease?

d. If a man tests positive on two successive tests, what is the probability that he has the disease?

Step 1 of 4 :

We have events A.

Let the person actually has disease it is denoted by A.

Let the test gives positive signal we denoted by +.

We know that

P(A)=0.005,

The probability of positive is

and

Then is

The probability of negative is

and

Our goal is to find :

a). If a man tests negative then we have to find the probability that he actually has the

disease.

b). For many medical tests, it is standard procedure to repeat the test when a positive signal,If

repeated tests are independent.We have to find the probability that a man will test

positive on two successive tests if the he has disease.

c). Assuming repeated tests are independent, we have to find the probability a man tests positive

on two successive tests if he does not have the disease.

d). If a man tests positive on two successive tests, we have to find the probability that he has the

disease.

a).

Given if a man tests negative then we have to find the probability that he actually has the disease.

Now we have to find .

The formula of the .

Substitute all the values.

0.000050750.0000508

0.0000508

Therefore the probability that he actually has the disease 0.0000508.

Step 2 of 4 :

b).

When a positive signal,If repeated tests are independent.

Now we have to find the probability that a man will test positive on two successive tests if the he has disease.

We need to find the probability of (positive ) ++/A .

The probability of positive is

Then the P(++/A) is

P(++/A) =

P(++/A) =

P(++/A) =

P(++/A) =

Therefore the probability that a man will test positive on two successive tests if the he has disease is 0.9801.