﻿ 4. In July of 2011 the state of Florida started testing all welfare recipients for the use of illeg
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### Question

4. In July of 2011 the state of Florida started testing all welfare recipients for the use of illegal drugs. Statistics suggest that 8 percent of adult Floridians use illegal drugs; let us assume that this is true for welfare recipients as well. Imagine that the drug test is 92 percent accurate, meaning that it gives the correct response in 92 out of 100 cases. (a) What is the probability that a randomly selected welfare recipient in Florida uses (b) What is the probability that a randomly selected welfare recipient in Florida does (c) What is the probability that a randomly selected welfare recipient in Florida has (d) Given that a randomly selected welfare recipient in Florida has a positive test, illegal drugs and has a positive test? not use illegal drugs but nevertheless has a positive test? a positive test? what is the probability that they use illegal drugs?

### Solution

xxxx we xxx xxxxx xxxx: P( xxxxxxx drugs ) = x.xx, therefore x( no xxxxx ) = 1 - 0.08 = x.xx xxxx, we xxx given xxxx xx xxxxx 92% xxxxxxx response. xxxxxxxxx, xx xxx:

P( xxxxxxxx | xxxxxxx xxxxx ) = x( negative | xx xxxxx ) = 0.92

x) xxx xxx probability xxxx a xxxxxxxy xxxxxxxx xxxxxxx uses xxxxxxx drugs xxx xxx x positive xxxx is xxxxxxxx xx: x( illegal xxxxx )P(positive | xxxxxxx xxxxx ) = 0.92*0.08 = x.xxxx xxxxxxxxx 0.0736 xx the xxxxxxxx xxxxxxxxxxy xxxx.

b) xxx the xxxxxxxxxxy xxxx x randomly xxxxxxxx do xxx xxx xxxxxxx drugs xxx still xxxx x xxxxxxxx test xx computed xx: x( xx drugs , positive ) = x.xx*(x - x.xx) = x.xxxx xxxxxxxxx x.xxxx is xxx required xxxxxxxxxxy xxxx. x) Now xxxxxxxxxxy that x xxxxxxxy xxxxxxxx person xxx a xxxxxxxx xxxx xx computed xx:

= x( xx xxxxx , xxxxxxxx ) + x( xxxxxxx drugs , positive ) = x.xxxx + x.xxxx

= x.xxxx xxxxxxxxx x.xxxx is xxx required xxxxxxxxxxy xxxx. x) Given xxxx there xx xxxxxxxx xxxx, probability xxxx they xxx xxxxxxx xxxxx is xxxxxxxx as:

= x( xxxxxxx drugs , positive ) / x ( xxxxxxxx )

= x.xxxx / 0.1472

= 0.5

xxxxxxxxx x.x xx the xxxxxxxx probability xxxx.