*72*

### Exercise

In a scientific article published in the *Journal of Public Health Statistics*, the authors estimated that in the general population, without any known risk factors, the probability that an individual is a carrier of a rare virus (let’s call it Virus X) is 0.00001. That is, in a population of 10 million people, there are approximately 100 carriers. To detect the virus in the blood, a diagnostic test is used. The characteristics of this test are:

\[

P(\text{Test Positive} \mid \text{Person is a Carrier}) = 0.998

\]

\[

P(\text{Test Negative} \mid \text{Person is NOT a Carrier}) = 0.9995

\]

Calculate the following:

1. The probability that a person is a carrier, given that the test result is positive.

2. The probability that a person is a carrier, given that the test result is negative.

3. The probability that a person is not a carrier, given that the test result is positive.

### Solution

Let \( C \) be the event that a person is a carrier, and \( \bar{C} \) be the event that a person is not a carrier. Let \( T^+ \) be the event that the test is positive, and \( T^- \) be the event that the test is negative.

#### Given:

\[

P(C) = 0.00001, \quad P(\bar{C}) = 1 – P(C) = 0.99999

\]

\[

P(T^+ \mid C) = 0.998, \quad P(T^- \mid \bar{C}) = 0.9995

\]

We need to find the following probabilities:

#### Step 1:

##### Probability that a person is a carrier given that the test result is positive (\(P(C \mid T^+)\)).

Using Bayes’ theorem:

\[

P(C \mid T^+) = \frac{P(T^+ \mid C) \cdot P(C)}{P(T^+)}

\]

Where \( P(T^+) \) is the total probability of a positive test result:

\[

P(T^+) = P(T^+ \mid C) \cdot P(C) + P(T^+ \mid \bar{C}) \cdot P(\bar{C})

\]

First, we need to calculate \( P(T^+ \mid \bar{C}) \), which is:

\[

P(T^+ \mid \bar{C}) = 1 – P(T^- \mid \bar{C}) = 1 – 0.9995 = 0.0005

\]

Now, substitute the values into the equation for \( P(T^+) \):

\[

P(T^+) = (0.998 \times 0.00001) + (0.0005 \times 0.99999)

\]

\[

P(T^+) = 0.00000998 + 0.000499995 = 0.000509975

\]

Finally, calculate \( P(C \mid T^+) \):

\[

P(C \mid T^+) = \frac{0.998 \times 0.00001}{0.000509975} \approx 0.01957

\]

So, the probability that a person is a carrier given that the test result is positive is approximately **0.01957**.

#### Step 2:

##### Probability that a person is a carrier given that the test result is negative (\(P(C \mid T^-)\)).

Using Bayes’ theorem:

\[

P(C \mid T^-) = \frac{P(T^- \mid C) \cdot P(C)}{P(T^-)}

\]

Where \( P(T^-) \) is the total probability of a negative test result:

\[

P(T^-) = P(T^- \mid C) \cdot P(C) + P(T^- \mid \bar{C}) \cdot P(\bar{C})

\]

Since \( P(T^- \mid C) = 1 – P(T^+ \mid C) = 1 – 0.998 = 0.002 \), we substitute:

\[

P(T^-) = (0.002 \times 0.00001) + (0.9995 \times 0.99999)

\]

\[

P(T^-) = 0.00000002 + 0.999490005 = 0.999490025

\]

Finally, calculate \( P(C \mid T^-) \):

\[

P(C \mid T^-) = \frac{0.002 \times 0.00001}{0.999490025} \approx 0.00000002

\]

So, the probability that a person is a carrier given that the test result is negative is approximately **0.00000002**.

#### Step 3:

##### Probability that a person is not a carrier given that the test result is positive (\(P(\bar{C} \mid T^+)\)).

Again, using Bayes’ theorem:

\[

P(\bar{C} \mid T^+) = \frac{P(T^+ \mid \bar{C}) \cdot P(\bar{C})}{P(T^+)}

\]

Substituting the values:

\[

P(\bar{C} \mid T^+) = \frac{0.0005 \times 0.99999}{0.000509975} \approx 0.98043

\]

So, the probability that a person is not a carrier given that the test result is positive is approximately **0.98043**.

### Summary of Answers:

1. Probability that a person is a carrier given that the test result is positive: **0.01957**

2. Probability that a person is a carrier given that the test result is negative: **0.00000002**

3. Probability that a person is not a carrier given that the test result is positive: **0.98043**