Math Problem Statement

A certain virus infects one in every 400 people. A test used to detect the virus in a person is positive 90% of the time if the person has the virus and 10% of the time if the person does not have the virus. Let A be the event "the person is infected" and B be the event "the person tests positive." Find the probability that a person has the virus given that they have tested positive.

Solution

We are asked to find the probability that a person has the virus given that they tested positive, P(AB)P(A | B). This is a conditional probability problem, which can be solved using Bayes' Theorem.

Bayes' Theorem is given by:

P(AB)=P(BA)P(A)P(B)P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)}

Where:

  • P(A)P(A) is the prior probability that the person is infected (event A).
  • P(BA)P(B | A) is the probability of testing positive given that the person has the virus.
  • P(B)P(B) is the total probability that a person tests positive.

Let's define the values:

  1. P(A)=1400P(A) = \frac{1}{400} (since one in every 400 people is infected).
  2. P(BA)=0.9P(B | A) = 0.9 (the test correctly detects the virus 90% of the time).
  3. P(BAc)=0.1P(B | A^c) = 0.1 (the test gives a false positive 10% of the time, where AcA^c is the event of not being infected).

Step 1: Calculate P(B)P(B)

To find P(B)P(B), the total probability of testing positive, we use the Law of Total Probability. It accounts for both cases when the person has the virus and when the person does not have the virus:

P(B)=P(BA)P(A)+P(BAc)P(Ac)P(B) = P(B | A) \cdot P(A) + P(B | A^c) \cdot P(A^c)

Where P(Ac)=1P(A)=399400P(A^c) = 1 - P(A) = \frac{399}{400}.

Now plug in the values:

P(B)=(0.9)1400+(0.1)399400P(B) = (0.9) \cdot \frac{1}{400} + (0.1) \cdot \frac{399}{400}

P(B)=0.9400+39.9400P(B) = \frac{0.9}{400} + \frac{39.9}{400}

P(B)=40.8400=0.102P(B) = \frac{40.8}{400} = 0.102

Step 2: Apply Bayes' Theorem

Now that we have P(B)P(B), we can apply Bayes' Theorem:

P(AB)=P(BA)P(A)P(B)=(0.9)14000.102P(A | B) = \frac{P(B | A) \cdot P(A)}{P(B)} = \frac{(0.9) \cdot \frac{1}{400}}{0.102}

P(AB)=0.9400×0.102=0.940.8P(A | B) = \frac{0.9}{400 \times 0.102} = \frac{0.9}{40.8}

P(AB)0.0221P(A | B) \approx 0.0221

Thus, the probability that a person has the virus given that they tested positive is approximately 2.21%.

Would you like any further clarification on this, or do you want to explore more details?


Here are 5 related questions you might find helpful:

  1. How does Bayes' Theorem apply in medical testing scenarios beyond this problem?
  2. How does increasing the accuracy of the test affect the probability P(AB)P(A | B)?
  3. What would happen if the false positive rate was reduced to 5% instead of 10%?
  4. How would the probability change if the infection rate increased to 1 in 200 people?
  5. How do false positives and false negatives impact medical decision-making?

Tip: Always check the assumptions and test accuracy when interpreting medical test results!

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Conditional Probability
Bayesian Inference

Formulas

Bayes' Theorem: P(A | B) = [P(B | A) * P(A)] / P(B)
Law of Total Probability: P(B) = P(B | A) * P(A) + P(B | A^c) * P(A^c)

Theorems

Bayes' Theorem
Law of Total Probability

Suitable Grade Level

Grades 11-12 or College Level