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Hoochie [10]
3 years ago
14

In the manufacturing of a chemical adhesive, 9% of all batches have raw materials from two different lots. This occurs when hold

ing 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. Round your answers to three decimal places (e.g. 98.765).
Mathematics
1 answer:
DochEvi [55]3 years ago
5 0

Answer:

a) P(A) = 0.09

b) P(A') = 0.91

c) P(B|A) = 0.4

d) P(B/A') = 0.05

e) P(A ∩ B) = 0.036

f) P(A ∩ B') = 0.054

g) P(B) = 0.0815

Step-by-step explanation:

Let A mean batch is made from two different lots and B mean batch requires additional processing

It is given in the question that

P(A) = 0.09

P(B|A) = 0.4

P(B|A') = 0.05

a) P(A) = 0.09 (given)

b) P(A') = 1-0.09 = 0.91

c) P(B|A) = 0.4 (given)

d) P(B|A') = 0.05 (given)

e) P(A ∩ B) = P(B|A) × P(A) = 0.4 × 0.09 = 0.036

f) P(A ∩ B') = P(A) - P(A ∩ B) = 0.09 - 0.036 = 0.054

g) P(B) = P(A ∩ B) + P(A' ∩ B)

P(B) = P(B|A) P(A) + P(B|A') P(A') = 0.4 × 0.09 + 0.05 × 0.91 = 0.0815

QED!

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Answer:

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Step-by-step explanation:

Let's start writing the sample space for this experiment :

S= { (1,1) , (1,2) , (1,3) , (1,4) , (1,5) , (1,6) , (2,1) , (2,2) , (2,3) , (2,4) , (2,5) , (2,6) , (3,1) , (3,2) , (3,3) , (3,4) , (3,5) , (3,6) , (4,1) , (4,2) , (4,3) , (4,4) , (4,5) , (4,6) , (5,1) , (5,2) , (5,3) , (5,4) , (5,5) , (5,6) , (6,1) , (6,2) , (6,3) , (6,4) , (6,5) , (6,6) }

Let's also define the event E ⇒

E : '' The sum of the two dice is 5 ''

We can describe the event by listing all the favorables cases from S ⇒

E = { (4,1) , (3,2) , (2,3) , (1,4) }

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P(E)=\frac{4}{36}=\frac{1}{9} ⇒

P(E)=\frac{1}{9}

Finally we are going to define the event F ⇒

F : '' The number of the first die is exactly 1 more than the number on the second die ''

⇒

F = { (2,1) , (3,2) , (4,3) , (5,4) , (6,5) }

Now given two events A and B ⇒

P ( A ∩ B ) = P(A,B)

We define the conditional probability as

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We calculate P(E)=\frac{1}{9} at the beginning of the question. We only need P(F,E).

Looking at the sets E and F we find that (3,2) is the unique result which is in both sets. Therefore is 1 result over the 36 possible results. ⇒

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Replacing both probabilities calculated in (I) :

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We find out that P(F|E)=\frac{1}{4}=0.25

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