Samples of a cast aluminum part are classified on the basis of surface finish (in microinches) and edge finish. The results of 100 parts are summarized as follows: Edge Finish
Excellent Good
Surface Excellent 80 2
Finish Good 10 8
Let A denote the event that a sample has excellent surface finish, and let B denote the event that a sample has excellent edge finish. If a part is selected at random, determine the following probabilities:
a) P(A)
b) P(B)
c) P(A')
d) P(A ∩ B)
e) P(A ∪ B)
f) P(A' ∪ B)

Step-by-step explanation:

From the question:

The given table is as follows:

                                              Edge Finish

                                          Excellent                  Good

Surface       Excellent           80                           2

Finish          Good                  10                            8

So, we are being told that,

A represent the event that a sample has an excellent surface finish, and let B denote the event that a sample has excellent edge finish

The objective is to fin the following probabilities

P(A)

A = (80 +2) = 82

The total samples of cast = 80 + 10 +2 + 8 = 100

∴ P(A) = 82/100

P(A) = 0.82

P(B)

B = 80+10

B = 90

P(B) = 90/100

P(B) = 0.90

c) P(A')

(A') are the sets  that are not in A but they are in the samples

i,e

(A') = 100 - 82

(A') = 18

So;

P(A')  = 100/18

P(A')  = 5.56

d)  P(A ∩ B)

A = ( 80, 2)

B = (80,10)

The intersection of A and B (i.e. A ∩ B) is the common value between them which is 80

P(A ∩ B) = 80/100

P(A ∩ B) = 0.80

e)  P(A ∪ B)

A = ( 80, 2)

B = (80,10)

The union of A and B is the addition of A and B

i.e. 80+2+10 = 92

P(A ∪ B) = 92/100

P(A ∪ B) =  0.92

f.  P(A' ∪ B)

A' = (10, 8)

B = (80,10)

The union of A complement and B is

(A' ∪ B) = 10 + 8 + 80

(A' ∪ B) = 98

P(A' ∪ B) =  98/100

P(A' ∪ B) = 0.98

b i think

step-by-step explanation:

not really sure