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10 votes
10 votes
3.8% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely accurate. 93.9% of those who have the disease test positive. However 4.1% of those who do not have the disease also test positive (false positives). A person is randomly selected and tested for the disease. What is the probability that the person has the disease given that the test result is positive? 0.475 0.038 0.525 0.905

User Manika
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1 Answer

18 votes
18 votes

ANSWER:

0.475

Explanation:

The probability of a person has disease given the test is positive:

P (disease) = 3.8% = 0.038

P (positive | disease) = 93.9% = 0.939

P (positive | no disease) = 4.1% = 0.041

P (no disease) = 100% - 3.8% = 96.2% = 0.962

The probability that the person has the disease given that the test result is positive is calculated as follows:


\begin{gathered} \text{ P\lparen infected \mid test positive\rparen }=\frac{\text{ P\lparen positive \mid infected\rparen }*\text{ \rbrack P \lparen infected\rparen}}{\text{ P \lparen positive\rparen}} \\ \\ \text{ P \lparen positive \mid infected\rparen }=\text{ P \lparen positive \mid disease\rparen = 0.939} \\ \\ \text{ P \lparen infected\rparen = P \lparen disease\rparen = 0.038} \\ \\ \text{ P \lparen positive\rparen = P \lparen positive \mid infected\rparen }*\text{ P \lparen infected\rparen }+\text{ P \lparen positive \mid no infected\rparen}*\text{ P \lparen no infected\rparen } \\ \\ \text{ P \lparen positive \mid infected\rparen =P \lparen positive \mid no disease\rparen = 0.041} \\ \\ \text{ P \lparen no infected\rparen = P \lparen no disease\rparen = 0.962} \\ \\ \text{ We replacing:} \\ \\ \text{ P \lparen positive\rparen = }0.038\cdot0.939+0.041\cdot0.962=0.075124 \\ \\ \text{ P\lparen infected \mid test positive\rparen }=(0.038\cdot0.939)/(0.075124) \\ \\ \text{ P\lparen infected \mid test positive\rparen = }\:0.47497=0.475 \end{gathered}

The correct answer is the first option: 0.475

User Ricky Boy
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