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OpenStudy (anonymous):

A 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: 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)? What is the probability that a person chosen at random will test positive but not have the disease (i.e., get a false positive)?

OpenStudy (anonymous):

@SolomonZelman how about dis 1?

OpenStudy (anonymous):

@DanJS halp me plz

OpenStudy (misty1212):

if this gets confusing, pick a number of people and work with that number make it a large power of ten

OpenStudy (misty1212):

say 10,000 people how many have the disease?

OpenStudy (anonymous):

100?

OpenStudy (misty1212):

ten out of a hundred is ten percent

OpenStudy (anonymous):

so 1000

OpenStudy (misty1212):

ok so we know 1,000 have the disease how many do not?

OpenStudy (anonymous):

5% of those?

OpenStudy (danjs):

sorry i'm late,, misty gots it for ya..

OpenStudy (anonymous):

it's ok

OpenStudy (misty1212):

you are thinking too hard there are 10,000 total (we made that up) 1,000 have the disease how many people don't have it?

OpenStudy (anonymous):

9000

OpenStudy (misty1212):

that makes more sense now lets compute some stuff

OpenStudy (misty1212):

of the 1,000 that have it, how many test positive?

OpenStudy (anonymous):

95%

OpenStudy (misty1212):

yes, 95% of them the question is not "what percent" of them but "how many" of them

OpenStudy (misty1212):

i.e. what is 95% of 1,000?

OpenStudy (anonymous):

950

OpenStudy (misty1212):

right exactly so now we are almost done with the first question out of the 10,000 people we know that 950 of them have the disease and also test positive making the probability the ratio \(\frac{950}{10,000}\)

OpenStudy (anonymous):

9.5%?

OpenStudy (misty1212):

that would be \(.095\) or \(9.5\%\) we could have arrived at this answer by simply computing \[0.1\times .95\] as well

OpenStudy (misty1212):

the second one is a bit harder

OpenStudy (misty1212):

well not really how many people do not have the disease? we already found that, it is 9,000 out of those how many test positive?

OpenStudy (anonymous):

95% or 8550

OpenStudy (misty1212):

lets back up and go slow

OpenStudy (misty1212):

there are 9,000 people who do NOT have the disease right?

OpenStudy (anonymous):

uhh 450

OpenStudy (misty1212):

this the question states It also has a 97% chance to indicate that the person does NOT have the condition if they really don't have it.

OpenStudy (misty1212):

if it is right 97% of the time, it must be wrong what percent of the time?

OpenStudy (anonymous):

is it 270

OpenStudy (misty1212):

yes 3% of 9,000 is 270

OpenStudy (misty1212):

so the answer to the second part, the probability it is a false positive, is \(\frac{270}{10,000}\)

OpenStudy (anonymous):

so a is 9.5% and b is 2.7%?

OpenStudy (misty1212):

yes it is

OpenStudy (anonymous):

ok thank you, i have another problem after this, it is the last one

OpenStudy (misty1212):

we could have got B by computing \[0.03\times 09\] also

OpenStudy (misty1212):

but it makes more sense i think if you do it with numbers

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