In the manufacturing of a chemical adhesive, 9% 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. Round your answers to three decimal places (e. g. 98.765).

Question

Mathematics

Mariela

2 February, 11:28

In the manufacturing of a chemical adhesive, 9% 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. Round your answers to three decimal places (e. g. 98.765).

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Answers (1)

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Gideon Hendrix · Answer 1

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!