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natetheman7740
09.09.2019 •
Mathematics
In the manufacturing of a chemical adhesive, 3% of all batches have raw materials from two different lots. this occurs when holding tanks are replenished and the remaining portion of a lot is insufficient to fill the tanks. only 5% of batches with material from a single lot require reprocessing. however, the viscosity of batches consisting of two or more lots of material is more difficult to control, and 40% of such batches require additional processing to achieve the required viscosity. let a denote the event that a batch is formed from two different lots, and let b denote the event that a lot requires additional processing. determine the following probabilities:
a. p(a)
b. p(a')
c. p(b\a)
d. p(b\a')
e. p(a ∩ b)
f. p(a ∩ b')
g. p(b)
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Ответ:
Step-by-step explanation:
By the problem we know that our events are:
So, according to that:
a)![P(A)=3%=0.03](/tpl/images/0226/1863/5c3a1.png)
Because P(A) and
are complementary events
b)![P(A^{'} )=1-P(A)=1-0.03=0.97=97%](/tpl/images/0226/1863/1fb7d.png)
Because of the problem, we know that:
c)![P(B/A)=0.4=40%](/tpl/images/0226/1863/fa224.png)
and,
d)![P(B/A^{'} )=0.05=5%](/tpl/images/0226/1863/b0677.png)
From the formula
e) P(A ∩ B)= P(A)*P(B/A)=![(0.03)*(0.4)=0.012](/tpl/images/0226/1863/7e3b2.png)
f) P(A ∩ B')=P(A)-P(A ∩ B)=![0.03-0.012=0.018](/tpl/images/0226/1863/174d6.png)
And, finally
g) P(B)=P(B/A)*P(A)+P(B/A')*P(A')=![(0.4)*(0.03)+(0.05)(0.97)=0.0605](/tpl/images/0226/1863/f071d.png)
Ответ: