SOLUTION: If a test is carried out and it identifies a disease in 95% of people who have it. It correctly identifies no disease in 94% of people who do not have it. In the population, 3% of
Algebra.Com
Question 1088334: If a test is carried out and it identifies a disease in 95% of people who have it. It correctly identifies no disease in 94% of people who do not have it. In the population, 3% of the people have the disease. What is the probability that you have the disease if you tested positive?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
D = person has disease
D' = person does not have disease
T = person tests positive
T' = person does not test positive (person tests negative)
----------------------------------------------------------
Let's read through the given text to see what we can pull out
Given: "If a test is carried out and it identifies a disease in 95% of people who have it"
Translation: P(T|D) = 0.95 .... call this Equation (1)
----------------
Given: "It correctly identifies no disease in 94% of people who do not have it"
Translation: P(T'|D') = 0.94 .... call this Equation (2)
----------------
Given: "In the population, 3% of the people have the disease"
Translation: P(D) = 0.03 .... call this Equation (3)
----------------
P(D)+P(D') = 1
P(D') = 1-P(D)
P(D') = 1-0.03 ... plug in Equation (3)
P(D') = 0.97 .... call this Equation (4)
----------------------------------------------------------
Question: "What is the probability that you have the disease if you tested positive?"
What we want to find: P(D|T)
----------------------------------------------------------
Use the conditional definition to say
P(T|D) = P(T and D)/P(D)
P(D)*P(T|D) = P(T and D)
P(T and D) = P(D)*P(T|D)
P(T and D) = 0.03*0.95 ... plug in Equation (3) and Equation (1)
P(T and D) = 0.0285 .... call this Equation (5)
Now use the fact that P(T|D') and P(T'|D') are complementary to say
P(T|D') + P(T'|D') = 1
P(T|D') = 1 - P(T'|D')
P(T|D') = 1 - 0.94 ... plug in Equation (2)
P(T|D') = 0.06 .... call this Equation (6)
We can also break down P(T|D') using the conditional definition
P(T|D') = P(T and D')/P(D')
P(D')*P(T|D') = P(T and D')
P(T and D') = P(D')*P(T|D')
P(T and D') = 0.97*0.06 ... plug in Equation (4) and Equation (6)
P(T and D') = 0.0582 ... call this Equation (7)
Law of Total Probability
P(T) = P(T and D) + P(T and D')
P(T) = 0.0285 + 0.0582 ... plug in Equation (5) and Equation (7)
P(T) = 0.0867 .... call this Equation (8)
Now onto the probability we want to find.
Use the conditional formula to get
P(D|T) = P(D and T)/P(T)
P(D|T) = P(T and D)/P(T)
P(D|T) = 0.0285/0.0867 ... plug in Equation (5) and Equation (8)
P(D|T) = 0.3287197231834
P(D|T) = 0.3287
Rounded to four decimal places, the probability in decimal form is 0.3287 which converts to 32.87% (both of these values are approximate)
RELATED QUESTIONS
A certain disease has an incidence rate of 0.7%. (This is percent of people in the... (answered by jorel1380)
A certain disease has an incidence rate of 0.6%. (This is percent of people in the... (answered by jorel1380)
The test accurately indicates the disease in 98% of the people who have it, and it... (answered by math_tutor2020)
One in a thousand people is a carrier of a particular heart disease. There is a test to... (answered by ewatrrr)
In a region 4% of the population is thought to have a certain disease. A standard... (answered by stanbon)
A new medical test has been designed to detect the presence of the mysterious... (answered by Boreal)
A new medical test has been designed to detect the presence of the mysterious... (answered by Boreal)
Suppose there is a test for a disease that correctly gives positive results for 95% of... (answered by kev82)
A medical test has been designed to detect the presence of a certain disease. Among those (answered by CPhill)