MATH SOLVE

2 months ago

Q:
# PLEASE HELP. WILL MARK BRAINLIESTA medical test has a 95% accuracy of detecting a Condition Z if the person has it. It also has a 97% chance to indicate that the person does not have the condition if they really don't have it. If the incidence rate of this disease is 10 out of every 100:1.] What is the probability that a person chosen at random will both test positive and actually have the disease (i.e., get a true positive)?2.] What is the probability that a person chosen at random will test positive but not have the disease (i.e., get a false positive)?

Accepted Solution

A:

Problems like this are complex and the guide to solve them is to always think of all pairs of possibilities; These are 4 now, namely (infected, positive result), (infected, negative result), (not infected, positive result), (not infected, negative result).

If your test shows positive, then either you are infected and the test is correct or you are not and the test is incorrect. The chance for the first event is 10%*95% by the multiplicative law of probabilities, or 0.095. The chance for the second one (to test positively while not infected) is: (1-0.97)* 0.90=0.027

This is the probability that a person is not infected, times the probability that the test indicates that he is infected. So, the total chance of these two scenarios is 0.027+0.095=0.122. The chance that a person with a positive result is actually positive is thus 0.095/0.122=77.9%. Hence, far from sure.

Similarly as above, the false positive is the complimentary probability. The probability across the population that you test positive while not having the disease is 0.027. Thus, the probability is 0.027/0.122=22.1%. This can be also calculated by 100%-77.9%=22,1%, so this serves as a test of our correctness.

If your test shows positive, then either you are infected and the test is correct or you are not and the test is incorrect. The chance for the first event is 10%*95% by the multiplicative law of probabilities, or 0.095. The chance for the second one (to test positively while not infected) is: (1-0.97)* 0.90=0.027

This is the probability that a person is not infected, times the probability that the test indicates that he is infected. So, the total chance of these two scenarios is 0.027+0.095=0.122. The chance that a person with a positive result is actually positive is thus 0.095/0.122=77.9%. Hence, far from sure.

Similarly as above, the false positive is the complimentary probability. The probability across the population that you test positive while not having the disease is 0.027. Thus, the probability is 0.027/0.122=22.1%. This can be also calculated by 100%-77.9%=22,1%, so this serves as a test of our correctness.