Math Problem Statement

A certain virus infects one in every 300 people. A test used to detect the virus in a person is positive 80% of the time if the person has the virus and 5% of the time if the person does not have the virus. (This 5% result is called a false positive.) Let A be the event "the person is infected" and B be the event "the person tests positive".

(a) Find the probability that a person has the virus given that they have tested positive, i.e. find P(Has Virus | Tested Positive).

Round your answer to the nearest tenth of a percent.

P(Has Virus | Tested Positive)= %

Solution

To solve for P(Has Virus | Tested Positive)P(\text{Has Virus | Tested Positive}), we can use Bayes' Theorem, which states:

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

where:

  • P(A)P(\text{A}) is the probability that a person has the virus.
  • P(B | A)P(\text{B | A}) is the probability that the test is positive given that the person has the virus.
  • P(B)P(\text{B}) is the probability that the test is positive.

Given values:

  1. The probability that a person has the virus, P(A)=1300P(\text{A}) = \frac{1}{300}.
  2. The probability that a test is positive given the person has the virus, P(B | A)=0.8P(\text{B | A}) = 0.8.
  3. The probability of a false positive, P(B | Not A)=0.05P(\text{B | Not A}) = 0.05.

We first need to calculate P(B)P(\text{B}), the probability that a person tests positive. This can happen in two ways:

  1. The person has the virus and tests positive.
  2. The person does not have the virus but still tests positive (false positive).

Thus,

P(B)=P(B | A)P(A)+P(B | Not A)P(Not A)P(\text{B}) = P(\text{B | A}) \cdot P(\text{A}) + P(\text{B | Not A}) \cdot P(\text{Not A})

where:

  • P(Not A)=1P(A)P(\text{Not A}) = 1 - P(\text{A}).

Step-by-Step Calculation

  1. Calculate P(A)P(\text{A}):

    P(A)=13000.00333P(\text{A}) = \frac{1}{300} \approx 0.00333

  2. Calculate P(Not A)P(\text{Not A}):

    P(Not A)=10.00333=0.99667P(\text{Not A}) = 1 - 0.00333 = 0.99667

  3. Calculate P(B)P(\text{B}):

    P(B)=(0.8×0.00333)+(0.05×0.99667)P(\text{B}) = (0.8 \times 0.00333) + (0.05 \times 0.99667) P(B)=0.002664+0.0498335=0.0524975P(\text{B}) = 0.002664 + 0.0498335 = 0.0524975

  4. Apply Bayes' Theorem to find P(A | B)P(\text{A | B}):

    P(A | B)=P(B | A)P(A)P(B)P(\text{A | B}) = \frac{P(\text{B | A}) \cdot P(\text{A})}{P(\text{B})} P(A | B)=0.8×0.003330.0524975P(\text{A | B}) = \frac{0.8 \times 0.00333}{0.0524975} P(A | B)0.0026640.05249750.0508P(\text{A | B}) \approx \frac{0.002664}{0.0524975} \approx 0.0508

  5. Convert to a percentage and round to the nearest tenth of a percent:

    P(A | B)5.1%P(\text{A | B}) \approx 5.1\%

Final Answer:

P(Has Virus | Tested Positive)=5.1%P(\text{Has Virus | Tested Positive}) = 5.1\%

Would you like further details on any steps, or do you have questions?


Here are five related questions:

  1. What is the probability of a false positive in this test scenario?
  2. How does a higher accuracy in the test affect the conditional probability?
  3. If the infection rate increases to 1 in 100, how does P(Has Virus | Tested Positive)P(\text{Has Virus | Tested Positive}) change?
  4. How can false negatives impact the test results?
  5. What is Bayes' Theorem, and where else is it used?

Tip: When interpreting conditional probabilities, remember that a low prevalence rate can significantly affect the predictive value of a positive test result.

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Math Problem Analysis

Mathematical Concepts

Probability
Conditional Probability
Bayes' Theorem

Formulas

Bayes' Theorem: P(A | B) = (P(B | A) * P(A)) / P(B)
Probability of an event not occurring: P(Not A) = 1 - P(A)
Total probability: P(B) = P(B | A) * P(A) + P(B | Not A) * P(Not A)

Theorems

Bayes' Theorem

Suitable Grade Level

Grades 10-12