Question

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)

Answer 1

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) = $\frac{P(F) * P(A/F)}{P(F) * P(A/F) + P(NF) *P(A/NF)}$ = $\frac{0.69*0.77}{0.69*0.77+0.31*0.20}$ = $\frac{0.5313}{0.5933}$

               = 0.8955