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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)

Answer :

amhugueth

Step-by-step explanation:

By the problem we know that our events are:

[tex]A=[/tex]A batch is formed from two different lots.

[tex]B=[/tex]A batch requires additional processing.

So, according to that:

a) [tex]P(A)=3%=0.03[/tex]

Because P(A) and [tex]P(A^{'} )[/tex] are complementary events

b) [tex]P(A^{'} )=1-P(A)=1-0.03=0.97=97%[/tex]

Because of the problem, we know that:

c) [tex]P(B/A)=0.4=40%[/tex]

and,

d) [tex]P(B/A^{'} )=0.05=5%[/tex]

From the formula

e) P(A ∩ B)= P(A)*P(B/A)=[tex](0.03)*(0.4)=0.012[/tex]

f) P(A ∩ B')=P(A)-P(A ∩ B)=[tex]0.03-0.012=0.018[/tex]

And, finally

g) P(B)=P(B/A)*P(A)+P(B/A')*P(A')=[tex](0.4)*(0.03)+(0.05)(0.97)=0.0605[/tex]

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