SOLUTION: The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags. Based on many test runs, the TSA determines that if a bag does

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Question 345439: The Transportation Safety Authority (TSA) has developed a new test to detect large amounts of liquid in luggage bags. Based on many test runs, the TSA determines that if a bag does contain large amounts of liquid, there is a probability of 0.92 the test will detect it. If a bag does not contain large amounts of liquid, there is a 0.1 probability the test will conclude that it does (a false positive). Suppose that in reality only 3 in 100 bags actually contain large amounts of liquid.
A)What is the probability a randomly selected bag will have a positive test? Give your answer to four decimal places.

Found 2 solutions by haileytucki, jrfrunner:
Answer by haileytucki(390) About Me  (Show Source):
You can put this solution on YOUR website!
q1
P[ +ve ] = P[ large & + ve ] + P[ small & +ve ]
= 2/100 x 98/100 + 98/100 x 10/100
= [196 + 980] /10^4
=1176/10^4
=0.1176
q2
P[ large | +ve ] = P[large & +ve] / P[+ve]
= 196/1176
=0.1667
q3
P[ smll | +ve ] = P[small & +ve] / P[+ve]
= 980/1176
= 0.8333

Answer by jrfrunner(365) About Me  (Show Source):
You can put this solution on YOUR website!
let D=event of detecting large amount of liquid or positive test
let L=event of bag containing large amounts of liquid
let S=event of bag containing small amounts of liquid
--
given
P(D/L)=0.92
P(D/S)=0.1
P(L)=3/100=0.03
Need to find P(D)
---
P(S)=1-P(L)=1-0.03-0.97
--
P(D)=P(D and L)+P(D and S)
=P(D/L)*P(L)+P(D/S)*P(S)
=0.92*0.03+0.1*0.97
=0.0276+0.097=0.1246