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The data shows the number of people who have and do not have Hepatitis C.

Further, the number of subjects who tested positive and negative is also given.

The following is the expression to compute the conditional probability of B given A.

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

It defines the occurrence of B with reference to a prior event A.

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 positive is equal to 337.

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

$P\left(C\right)=\frac{337}{1500}$

The number of subjects who have Hepatitis C and tested positive is equal to 335.

The probability of selecting a subject who tested positive and has Hepatitis C is given by:

$P\left(AandC\right)=\frac{335}{1500}$

The probability of selecting a subject who has Hepatitis C, given that he/she tested positive, is calculated as follows:

$$\begin{gathered} P\left(A|C\right)=\frac{P\left(AandC\right)}{P\left(C\right)} \\ =\frac{\frac{335}{1500}}{\frac{337}{1500}} \\ =\frac{335}{337} \\ =0.9941 \end{gathered}$$

Therefore, the probability of selecting a subject who has Hepatitis C, given that he/she tested positive, is equal to 0.9941.

The event of having Hepatitis C, given that the test result is positive, indicates the effectiveness of the test. As the probability value is approximately equal to 1, it can be said that the test is highly effective in detecting the disease.