Disease Test Probability Calculation

This problem involves calculating the probability of a person actually having a disease given a positive diagnostic test result. This is a classic application of Bayes' theorem.

Understanding the Given Information

• Total population: Let's assume a large population.
• Percentage of population with the disease: 1%.
• Percentage of population without the disease: 99%.
• Test reliability (Sensitivity): 98% of sick people test positive.
• Test reliability (Specificity): 98% of healthy people test negative.

Defining Events and Probabilities

Let's define the events:

• D: The person has the disease.
• ND: The person does not have the disease (is healthy).
• P: The person tests positive.
• N: The person tests negative.

From the problem statement, we have the following probabilities:

• Probability of having the disease, \(P(D) = 1\% = 0.01\).
• Probability of not having the disease, \(P(ND) = 1 - P(D) = 1 - 0.01 = 0.99\).
• Probability of testing positive given the person has the disease (Sensitivity), \(P(P|D) = 98\% = 0.98\).
• Probability of testing negative given the person does not have the disease (Specificity), \(P(N|ND) = 98\% = 0.98\).

We need to find the probability that the person actually has the disease given that they tested positive, which is \(P(D|P)\).

Calculating Necessary Probabilities

We are given \(P(N|ND) = 0.98\). The probability of testing positive given the person does not have the disease (False Positive Rate) is:

\(P(P|ND) = 1 - P(N|ND) = 1 - 0.98 = 0.02\).

Applying Bayes' Theorem

Bayes' theorem states that:

\(P(D|P) = \frac{P(P|D) \times P(D)}{P(P)}\)

To use this formula, we first need to find the overall probability of testing positive, \(P(P)\). A person can test positive in two ways:

1. They have the disease AND test positive (\(D \text{ and } P\)).
2. They do not have the disease AND test positive (\(ND \text{ and } P\)).

So, the total probability of testing positive is:

\(P(P) = P(P|D) \times P(D) + P(P|ND) \times P(ND)\)

Substitute the known values:

\(P(P) = (0.98 \times 0.01) + (0.02 \times 0.99)\)

\(P(P) = 0.0098 + 0.0198\)

\(P(P) = 0.0296\)

Final Calculation

Now we can calculate \(P(D|P)\) using Bayes' theorem:

\(P(D|P) = \frac{P(P|D) \times P(D)}{P(P)}\)

\(P(D|P) = \frac{0.98 \times 0.01}{0.0296}\)

\(P(D|P) = \frac{0.0098}{0.0296}\)

To convert this to a percentage, multiply by 100:

\(P(D|P) \approx 0.33108\)

\(P(D|P) \approx 33.108\%\)

Conclusion

The probability that the person actually has the disease given a positive test result is approximately 33.1%. Comparing this to the given options, the closest value is 33%.

This result shows that even with a seemingly reliable test (98%), when the prevalence of the disease in the population is low (1%), a positive test result does not guarantee a high probability of actually having the disease. A significant portion of positive results in this scenario come from healthy people (false positives).