Alaboratory blood test is 99% effective in detecting a certain disease when it is in fact, present. however, the test also yields a false positive result for 0.5% of the healthy person tested (i. e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). if 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive ?
a. 0.1654
b. 0.1982
c. 0.1827
d. 0.1346

A. 0.1654

Step-by-step explanation:

Let's call :

D: The event in which a person has a disease

ND: The event in which a person doesn't have a disease

TD: The event in which the test result is positive

Therefore, the probability that a person has the disease given that his test result is positive is calculated as:

P(D/TD) =P(D∩TD)/P(TD)

Where P(D∩TD) is the probability that a person has a disease and the test result is positive and P(TD) is the probability that the test result is positive.

So, The probability P(D∩TD) is calculated as a multiplication of:

P(D∩TD)=0.1% * 99% = 0.099%

Because 0.1% is the percentage of population that actually has the disease and 99% is the probability that the test detect the disease when it is present.

Then, for calculate the probability P(TD) we need to sum the probability of the following cases:

P(ND∩TD)= (100%-0.1%)*0.5%=0.4995%

Because (100%-0.1%) is the percentage of population that doesn't have the disease and 0.5% is the probability that the test imply that a person has the disease when it is not present.

So, P(TD) is calculated as:

P(TD) = P(D∩TD) + P(ND∩TD)

P(TD) = 0.099% + 0.4995%

P(TD) = 0.5985%

Finally, P(D/TD) is:

answer: the answer to this problem is yes because 4n + 4 does equal 2m - 6

step-by-step explanation: first you have to calculate 2m - 6 second you have to calculate 4n + 4 finally you have to get your answer