First arc (centered at B)
1 answer:
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Answer:
(1) 0.5933
(2) 0.8955
Step-by-step explanation:
We are given that all bags entering a research facility are screened.
Let Probability that bags entering the building contains forbidden material,
P(F) = 0.69
Probability that bags entering the building does not contains forbidden material, P(NF) = 1 - 0.69 = 0.31
Let event A = alarm gets triggered
Probability that alarm gets trigger given the bags contain forbidden material, P(A/F) = 0.77
Probability that alarm gets trigger given the bags does not contain forbidden material, P(A/NF) = 0.20
(1) Probability that a bag triggers the alarm, P(A) ;
P(A) = P(F) * P(A/F) + P(NF) * P(A/NF)
= (0.69 * 0.77) + (0.31 * 0.20) = 0.5313 + 0.062
= 0.5933
Therefore, probability that a bag triggers the alarm is 0.5933 .
(2) Probability that a bag that triggers the alarm will actually contain forbidden material is given by P(F/A) ;
Using Bayes' Theorem;
P(F/A) = =
=
= 0.8955