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One percent of all individuals in a certain population are carriers of a particular disease. A diagnostic test for this disease has a \(90 \%\) detection rate for carriers and a \(5 \%\) detection rate for noncarriers. Suppose the test is applied independently to two different blood samples from the same randomly selected individual. a. What is the probability that both tests yield the same result? b. If both tests are positive, what is the probability that the selected individual is a carrier?

Short Answer

Expert verified
a) 0.9055; b) 0.7666.

Step by step solution

01

Identify Key Probabilities

Let's define some notations and probabilities:- Let \(C\) denote that the individual is a carrier and \(eg C\) denote that the individual is not a carrier.- \(P(C) = 0.01\) and \(P(eg C) = 0.99\).- The probability that a test detects the disease when the individual is a carrier is \(P(+|C) = 0.90\).- The probability that a test falsely detects the disease in a non-carrier is \(P(+|eg C) = 0.05\).- Thus, \(P(-|C) = 0.10\) and \(P(-|eg C) = 0.95\).
02

Calculate Probability of Both Tests Yielding Same Result for a Carrier

For a carrier, both tests yield the same result if:1. Both tests are positive: \(P(+,+|C) = P(+|C)^2 = (0.90)^2 = 0.81\).2. Both tests are negative: \(P(-,-|C) = P(-|C)^2 = (0.10)^2 = 0.01\).
03

Calculate Probability of Both Tests Yielding Same Result for a Non-carrier

For a non-carrier, both tests yield the same result if:1. Both tests are positive: \(P(+,+|eg C) = P(+|eg C)^2 = (0.05)^2 = 0.0025\).2. Both tests are negative: \(P(-,-|eg C) = P(-|eg C)^2 = (0.95)^2 = 0.9025\).
04

Total Probability of Both Tests Giving Same Result

Combine the probabilities from Step 2 and Step 3 using the law of total probability:\[P\text{(Same Result)} = P(C) \cdot (P(+,+|C) + P(-,-|C)) + P(eg C) \cdot (P(+,+|eg C) + P(-,-|eg C))\] Substitute in the values calculated earlier:\[P\text{(Same Result)} = 0.01 \cdot (0.81 + 0.01) + 0.99 \cdot (0.0025 + 0.9025) = 0.0082 + 0.8973 = 0.9055\]
05

Calculate Probability of Both Tests Positive Given Carrier

We need \(P(C|+,+)\), found using Bayes' theorem:\[P(C|+,+) = \frac{P(+,+|C) \cdot P(C)}{P(+,+)}\]- \(P(+,+|C) = 0.81\) as calculated in Step 2.- \(P(+,+) = P(+,+|C) \cdot P(C) + P(+,+|eg C) \cdot P(eg C)\).- \(P(+,+) = 0.81 \cdot 0.01 + 0.0025 \cdot 0.99 = 0.0081 + 0.002475 = 0.010575\).\[P(C|+,+) = \frac{0.81 \cdot 0.01}{0.010575} = \frac{0.0081}{0.010575} \approx 0.7666\]
06

Conclusion

The probability that both tests yield the same result is approximately \(0.9055\). If both tests are positive, the probability that the individual is a carrier is approximately \(0.7666\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bayes' Theorem
Bayes' Theorem is a fundamental concept in probability used to update the probability of a hypothesis as more evidence becomes available. It is especially useful in contexts like diagnostic tests and disease detection. The formula for Bayes' Theorem is:
  • \(P(A|B) = \frac{P(B|A) \cdot P(A)}{P(B)}\)
Here, \(P(A|B)\) is the probability of event A occurring given that B is true. It's the 'posterior' probability. \(P(B|A)\) denotes likelihood, showing how likely B is if A is true. \(P(A)\) is the 'prior' probability of A before observing B. \(P(B)\) is the total probability of B occurring.

In the context of test results for disease carriers, Bayes' Theorem helps determine how likely it is that a person is a carrier if both tests return positive. It uses prior knowledge of disease prevalence and test accuracy to make this determination a lot more reliable.
Diagnostic Tests
Diagnostic tests are designed to detect a particular outcome or condition, such as being a carrier of a disease. Two key metrics for diagnostic tests are sensitivity and specificity.

Sensitivity is the test's ability to correctly identify those with the disease (true positive rate). In our problem, if the test has a 90% sensitivity, it means it accurately detects 90% of true carriers.

Specificity is the test's ability to correctly identify those without the disease (true negative rate). A specificity of 95% means that 95% of non-carriers test negative. Understanding these values helps us judge a diagnostic test's reliability and usefulness, which is crucial for accurate probability calculations.
Disease Detection
Disease detection involves identifying the presence or absence of a disease or condition in an individual. In our problem, the diagnostic test aims to detect a disease with an overall prevalence of 1% in the population.

When calculating the probability of being a carrier based on multiple test results, we consider both the accuracy of each test and the likelihood of different outcomes. For instance:
  • A high detection rate in carriers helps confirm true positive results.
  • A low false-positive rate amongst non-carriers helps reduce the risk of false positives.
Accurately detecting diseases is critical for treatment and management, making these calculations significant in real-world scenarios.
Probability Calculations
Probability calculations are central to understanding how likely an event is to occur. They help predict outcomes given specific conditions. In probability, events such as test outcomes are modeled to understand their likelihood.

For our exercise:
  • The probability that both tests yield the same result involves understanding combinations of events for both carriers and non-carriers.
  • Combining probabilities requires the use of multiplication for independent events and addition for mutually exclusive ones.
In the exercise, calculating the total probability that both tests give the same result considers events where both tests return the same outcome. This involves a logical combination of the probabilities for carriers and non-carriers, emphasizing the powerful use of probability in conjunction with Bayes' Theorem for insightful insights.

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