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.
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)

Step-by-step explanation:

Hello!

In a chemical adhesive manufacturing company you can encounter the following situations:

The adhesive is made in batches.

9% of all batches have raw materials from two different lots.

5% of all batches with material from a single lot require reprocessing.

40% of all batches with the material of two or more lots require additional processing.

You have two events:

A: A batch is formed with two different lots.

B: The lot requires additional processing.

And you can define their complements:

A': A batch is formed with one lot.

B': The lot does not require additional processing.

a. This probability is given in the text "9% of all batches have raw materials from two different lots. ", symbolically:

P(A)= 0.09

b. If the probability of a batch containing two lots is 9%, then it's complimentary event has a probability equal to:

P(A')= 1 - 0.09= 0.91

c. P(B/A) It is a conditional probability and you can interpret it as "the lot requires additional processing given that the batch is formed with two lots"

This information is also given:

P(B/A)= 0.40

d. P(B/A') Is the probability of the lot requiring additional processing given that the batch is formed with one lot.

P(B/A')= 0.05

e. Now you have to calculate the probability that "the batch is made with two lots" and "the lot requires additional processing"

A and B aren't independent events, which means that the occurrence of one of them modifies the probability of occurrence of the other one in two replications of the experiment.

P(A ∩ B)= P(A) * P(B/A)

P(A ∩ B)= 0.09 * 0.40= 0.036

f. You have to calculate the probability of "batch is formed with two different lots" and "the lot does not require additional processing", these two events aren't independent.

You cannot calculate the probability of this intersection the same way you did in item e. because we have no information about the probability of P(B'/A). But the P(A) is equal to the summation of P(A ∩ B) and P(A ∩ B'), so:

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

g. To calculate the probability of the event "The lot requires additional processing." you have to add P(A ∩ B)+P(A' ∩ B)

First calculate P(A' ∩ B) = P(A')* P(B/A')= 0.91*0.05= 0.0455

P(B)= P(A ∩ B) + P(A' ∩ B)= 0.036 + 0.0455= 0.0815

I hope you have a SUPER day!

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: