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)

(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

all it means is function 3

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)