In Exercises 17–20, refer to the accompanying table showing results from a Chembio test for hepatitis C among HIV-infected patients (based on data from a variety of sources).

	Positive Test Result	Negative Test Result
Hepatitis C	335	10
No Hepatitis C	2	1153

Negative Predictive Value Find the negative predictive value for the test. That is, find the probability that a subject does not have hepatitis C, given that the test yields a negative result. Does the result make the test appear to be effective?

The given data consists of the number of people with and without Hepatitis C categorized as per the test results; negative or positive.

Conditional probability is computed for an event whose occurrence is based on another event.

The conditional probability of B, given A, has the following notation:

$P\left(B|A\right)=\frac{P\left(AandB\right)}{P\left(A\right)}$

Here,$P\left(AandB\right)$ refers to the probability of events A and B happening together.

Let A be the event of selecting a subject who has Hepatitis C

Let B be the event of selecting a subject who does not have Hepatitis C.

Let C be the event of selecting a subject who has a positive test result.

Let D be the event of selecting a subject who has a negative test result.

The following table consists of the total frequency under each category:

The total number of subjects is equal to 1500.

The number of subjects who tested negative is equal to 1163.

The probability of selecting a subject who tested negative is given as follows:

$P\left(D\right)=\frac{1163}{1500}$

The number of subjects without Hepatitis C who have tested negative is equal to 1153.

The probability of selecting a subject who tested negative and does not have Hepatitis C is given by:

$P\left(BandD\right)=\frac{1153}{1500}$

The probability of selecting a subject who does not have Hepatitis C, given that he/she tested negative, is calculated as follows:

$$\begin{gathered} P\left(B|D\right)=\frac{P\left(BandD\right)}{P\left(D\right)} \\ =\frac{\frac{1153}{1500}}{\frac{1163}{1500}} \\ =\frac{1153}{1163} \\ =0.9914 \end{gathered}$$

Therefore, the probability of selecting a subject who does not have Hepatitis C, given that he/she tested negative, is equal to 0.9914.

The event of not having Hepatitis C, given that the test result is negative, indicates one aspect of the effectiveness of the test, which is correctly detecting the disease.

As the probability value is very high (close to one), it can be said that the test is effective in detecting the disease.