Respuesta :
Answer:
The probability that someone who tests positive has the genetic disease is approximately 0.166 or 16.6% who tests positive actually has the disease
Step-by-step explanation:
The number of persons in 10,000 people that have the disease = One
The percentage of people with the disease that test positive = 99.6%
The percentage of people who do not have the disease test positive = 0.05%
The probability that someone who tests positive has the genetic disease
[tex]P(disease | positive) = \dfrac{P(disease) \cdot P(positive | disease)}{P(disease) \cdot P(positive | disease) + P(nodisease) \cdot P(positive | nodisease)}[/tex]
P(disease) = 0.01% = 0.0001
P(positive disease) = 99.6% = 0.996
P(nodisease) = (10,000,000 - 100)/10,000,000 = 99.99% = 0.9999
P(positivenodisease) = 0.05% = 0.0005
Whereby 10,000,000 people are tested, 1000 out of the 1,000,000 will have the disease, 1,000 - 0.996 × 1000 = 4 people out of the 1,000 will test negative while 996 will test positive. From the 9,999,000 people who do not have the disease, 9,999,000 × 0.0005 = 4999.5 will give positive test results.
Therefore, the total number of people that tests positive = 4,999.5 + 996 = 5,995.5
Therefore, out of the 5,995.5 that test positive for the disease, 996 will test positive
The probability that someone who tests positive has the genetic disease, P(disease positive) = 996/5,995.5 ≈ 0.166 or 16.6%
Therefore, approximately 16.6% of the people that test positive for the disease actually has the disease
We have;
[tex]P(disease | positive) = \dfrac{0.0001 \times 0.996}{0.0001 \times 0.996 + 0.9999 \times 0.0005} = 0.16612459344[/tex]
