All bags entering a research facility are screened The scree

All bags entering a research facility are screened. The screening process is not perfect so that 77% of the bags that contain forbidden material trigger an alarm and 20% of the bags that do not contain forbidden material also trigger the alarm. If 69% of bags entering the building contains forbidden material,

(1) what is the probability that a bag triggers the alarm? (round your answer to 4 decimal places)

(2) what is the probability that a bag that triggers the alarm will actually contain forbidden material? (round your answer to 4 decimal places)

Solution

Let

T = triggers alarm
F = has forbidden material

Thus,

1)

P(T) = P(F) P(T|F) + P(F\') P(T|F\')

= 0.69*0.77 + (1-0.69)*0.20

= 0.5933 [ANSWER]

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2)

P(F|T) = P(F) P(T|F)/P(T)

= 0.69*0.77/0.5933

= 0.895499747 [ANSWER]