Disks of polycarbonate plastic from a supplier are analyzed for scratch and shock resistance. The results from 108 disks are summarized as follows: Shock Resistance Scratch Resistance High Low High 55 10 Low 13 30 Let A denote the event that a disk has high shock resistance, and let B denote the event that a disk has high scratch resistance. If a disk is selected at random, determine the following probabilities. P(A)=86/100P(B)=79/100P(A')=7/50P(A U B)=95/100P(A' U B)= ???

i) 0.63

ii) 0.37

iii) 0.37

iv) 0.77

v) 0.51

Step-by-step explanation:

Total resistance value = 108

A = high shock resistance

B = high scratch resistance

when a disk is selected at random

i) P(A) = P ( high shock resistance )

          = ( 55 + 13 ) / 108

          = 68/108 =  17/27   = 0.63

ii) P( B ) = P ( high scratch resistance )

            = ( 10 + 30 ) / 108

            = 40 / 108 = 10/27  = 0.37

iii) P(A') = 1 - P(A)

            = 1 - 17/27 = 10/27  = 0.37

iv) P( A U B ) = P(A) + P(B) - P( A n B )

                    = 0.63 + 0.37 - 0.233

                    = 0.766 ≈ 0.77

v) P( A' U B ) = P(A') + P(B) - P(A n B )

                     = 0.37 + 0.37 - 0.233

                     = 0.507 ≈ 0.51

the mean and mad

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