Suppose 1% of Americans have a particular disease. Of the Americans who actually
ID: 3271677 • Letter: S
Question
Suppose 1% of Americans have a particular disease. Of the Americans who actually have the disease, a diagnostic test produced by a pharmaceutical company gives a positive result 90% (f the time. Of the Americans who don't actually have the disease, the sane diagnostic test gives a positive result 2% of the time. This diagnostic test never produces an inconclusive results (i.e. it always comes back cither positive or negative). Define the events PO = {Positive test result} N = {Negative test result} D = {Disease is actually present}, and suppose a person is randomly chosen from the population. (a) What is the probability the person doesn't have the disease? (b) What is the probability the person tests negative given the person actually has the disease? (c) What is the probability the person tests negative given the person doesn't actually have the disease? (d) What is the probability the person tests positive? (e) What is the probability the person tests negative? (f) What is the probability the person actually has the disease given the person tests positive?Explanation / Answer
Ans:
P(D)=0.01,P(not Desease)=1-0.01=0.99
a)P(person does not have desease)=0.99
b)P(positive/actually have desease)=0.9
P(negative/actually have desease)=1-0.9=0.1
c)P(positive/does not actually have desease)=0.02
P(negative/does not actually have desease)=1-0.02=0.98
d)P(positive)=P(positive/actually have desease)*P(actually have desease)+P(positive/does not actually have desease)*P(not desease)
=0.9*0.01+0.02*0.99
=0.0288
e)
P(negative)=P(negative/actually have desease)*P(actually have desease)+P(negtaive/does not actually have desease)*P(not desease)
=0.1*0.01+0.98*0.99
=0.9712
f)P(actually have desease/positive)=P(positive/actually have desease)*P(actually have desease)/P(positive)
=(0.9*0.01)/0.0288
=0.009/0.0288
=0.3125
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