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KillerSteamcar
26.02.2020 •
Mathematics
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)
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Ответ:
(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{0.5313}{0.5933}](/tpl/images/0524/7136/82870.png)
= 0.8955
Ответ:
Rotate triangle A by 180° and then translate triangle A 5 units to the right. After this, Dilate it by a scale factor of "2"