Question

All bags entering a research facility are screened. Ninety-seven percent of the bags that contain forbidden material trigger an alarm. Fifteen percent of the bags that do not contain forbidden material also trigger the alarm. If 1 out of every 1,000bags entering the building contains forbidden material, what is the probabilitythat a bag that triggers the alarm will actually contain forbidden material?(a) 0.00097(b) 0.00640(c) 0.03000(d) 0.14550(e) 0.97000

Answer 1

Answer:

(b) 0.00640

Step-by-step explanation:

Let

P(A|F) = The probability of bags that contain forbidden material trigger an alarm = 0.97 or 97%

P(A| not F)= The probability of bags that do not contain forbidden material also trigger the alarm= 0.15 or 15%

P(F) = the probability that a bag contain forbidden material = 0.001

P(not F) = the probability that a bag does not contain forbidden material = 0.999

Then P(A)= P(F) × P(A|F)  + P(not F)  × P(A| not F) = 0.001×0.97 + 0.999×0.15 = 0.15082

Then the probability that a bag that triggers the alarm will actually contain forbidden material =$P(F|A)=\frac{P(F)*P(A|F)}{P(A)}$

=$\frac{0.001*0.97}{0.15064}$ ≈ 0.0064