a certain virus infects one in every 250 people. a test used to detect the virus in a person is positive 80% of the time when the person has the virus and 10% of the time when the person does not have the virus. (this 10% result is called a false positive.) let a be the event he person is infected\ and b be the event he person tests positive.\
(a) using bayes theorem, when a person tests positive, determine the probability that the person is infected.
(b) using bayes theorem, when a person tests positive, determine the probability that the person is not infected.

(a) the probability that a person is infected when a person tests positive is .
(do not round until final answer. then round to three decimal places as needed.)

Question

a certain virus infects one in every 250 people. a test used to detect the virus in a person is positive 80% of the time when the person has the virus and 10% of the time when the person does not have the virus. (this 10% result is called a false positive.) let a be the event 	he person is infected\ and b be the event 	he person tests positive.\
(a) using bayes theorem, when a person tests positive, determine the probability that the person is infected.
(b) using bayes theorem, when a person tests positive, determine the probability that the person is not infected.

(a) the probability that a person is infected when a person tests positive is  .
(do not round until final answer. then round to three decimal places as needed.)

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Bayes Theorem", "Posterior Probability" ], "new_concepts": [], "current_concepts": [ "Bayes Theorem", "Posterior Probability", "Conditional Probability Notation" ] } ## Identify the given probabilities Using the *Bayes Theorem* and *Posterior Probability* knowledge points $$ \begin{aligned} &P(A) = \frac{1}{250} = 0.004 \ &P(A') = 1 - P(A) = 1 - 0.004 = 0.996 \ &P(B \mid A) = 0.80 \ &P(B \mid A') = 0.10 \end{aligned} $$ ## Calculate the probability for part (a) Using the *Bayes Theorem* and *Posterior Probability* knowledge points $$ \begin{aligned} P(A \mid B) &= \frac{P(A) \cdot P(B \mid A)}{P(A) \cdot P(B \mid A) + P(A') \cdot P(B \mid A')} \ &= \frac{0.004 \cdot 0.80}{0.004 \cdot 0.80 + 0.996 \cdot 0.10} \ &= \frac{0.0032}{0.0032 + 0.0996} \ &= \frac{0.0032}{0.1028} \approx 0.031128 \approx 0.031 \end{aligned} $$ ## Calculate the probability for part (b) Using the *Bayes Theorem* and *Posterior Probability* knowledge points $$ \begin{aligned} &P(B' \mid A) = 1 - P(B \mid A) = 1 - 0.80 = 0.20 \ &P(B' \mid A') = 1 - P(B \mid A') = 1 - 0.10 = 0.90 \ &P(A' \mid B') = \frac{P(A') \cdot P(B' \mid A')}{P(A') \cdot P(B' \mid A') + P(A) \cdot P(B' \mid A)} \ &= \frac{0.996 \cdot 0.90}{0.996 \cdot 0.90 + 0.004 \cdot 0.20} \ &= \frac{0.8964}{0.8964 + 0.0008} \ &= \frac{0.8964}{0.8972} \approx 0.999108 \approx 0.999 \end{aligned} $$ ## Question 1 The probability that a person is infected when a person tests positive is 0.031. ## Question 2 The probability that a person is not infected when a person tests negative is 0.999. { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Statistics", "Bayes Theorem" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the given probabilities

Calculate the probability for part (a)

Calculate the probability for part (b)

Answer:

Question 1

Question 2