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)?
@SolomonZelman how about dis 1?
@DanJS halp me plz
if this gets confusing, pick a number of people and work with that number make it a large power of ten
say 10,000 people how many have the disease?
100?
ten out of a hundred is ten percent
so 1000
ok so we know 1,000 have the disease how many do not?
5% of those?
sorry i'm late,, misty gots it for ya..
it's ok
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?
9000
that makes more sense now lets compute some stuff
of the 1,000 that have it, how many test positive?
95%
yes, 95% of them the question is not "what percent" of them but "how many" of them
i.e. what is 95% of 1,000?
950
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 \(\frac{950}{10,000}\)
9.5%?
that would be \(.095\) or \(9.5\%\) we could have arrived at this answer by simply computing \[0.1\times .95\] as well
the second one is a bit harder
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?
95% or 8550
lets back up and go slow
there are 9,000 people who do NOT have the disease right?
uhh 450
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.
if it is right 97% of the time, it must be wrong what percent of the time?
is it 270
yes 3% of 9,000 is 270
so the answer to the second part, the probability it is a false positive, is \(\frac{270}{10,000}\)
so a is 9.5% and b is 2.7%?
yes it is
ok thank you, i have another problem after this, it is the last one
we could have got B by computing \[0.03\times 09\] also
but it makes more sense i think if you do it with numbers
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