Suppose that a test for opium use has a 2% false positiverate and a 5% false negative rate. That is, 2% of peoplewho do not use opium test positive for opium, and 5% ofopium users test negative for opium. Furthermore, sup-pose that 1% of people actually use opium. a)Find the probability that someone who tests negativefor opium use does not use opium. b)Find the probability that someone who tests positivefor opium use actually uses opium.

ion

step-by-step explanation:

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85.560%

step-by-step explanation:

conditional probability is defined as:

p(a|b) = p(a∩b) / p(b)

in other words, the probability that a occurs, given that b has occurred = the probability that both a and b occur / the probability that b occurs.

in this example:

p(turns 64 | turns 44) = p(turns 64 & turns 44) / p(turns 44).

if someone turns 64, they must have already turned 44, so:

p(turns 64 | turns 44) = p(turns 64) / p(turns 44)

on page 42, we see that of the original 100,000, 94944 turn 44, and 81234 turn 64.   therefore:

p(turns 64 | turns 44) = 81234 / 94944

p(turns 64 | turns 44) = 0.85560

what? that isnt a question

step-by-step explanation: