A recording company obtains the blank CDs used to produce its labels from three compact disk manufacturers: I, II, and III. The quality control department of the company has determined that 3% of the compact disks produced by manufacturer I are defective, 5% of those produced by manufacturer II are defective, and 5% of those produced by manufacturer III are defective. Manufacturers I, II, and III supply 36%, 54%, and 10%, respectively, of the compact disks used by the company. What is the probability that a randomly selected label produced by the company will contain a defective compact disk?

We are given that the quality control department of the company has determined that 3% of the compact disks produced by manufacturer I are defective, 5% of those produced by manufacturer II are defective, and 5% of those produced by manufacturer III are defective. Manufacturers I, II, and III supply 36%, 54%, and 10%, respectively, of the compact disks used by the company.

Let the Probability that Manufacturer I supply compact disks = P(M I) = 0.36

Probability that Manufacturer II supply compact disks = P(M II) = 0.54

Probability that Manufacturer III supply compact disks = P(M III) = 0.10

Also, let D = defective compact disks

Probability that compact disks produced by manufacturer I are defective = P(D/M I) = 0.03

Probability that compact disks produced by manufacturer I are defective = P(D/M II) = 0.05

Probability that compact disks produced by manufacturer I are defective = P(D/M III) = 0.05

Now, probability that a randomly selected label produced by the company will contain a defective compact disk is given by;

= P(M I) [tex]\times[/tex] P(D/M I) + P(M II) [tex]\times[/tex] P(D/M II) + P(M III) [tex]\times[/tex] P(D/M III)

= 0.36 [tex]\times[/tex] 0.03 + 0.54 [tex]\times[/tex] 0.05 + 0.10 [tex]\times[/tex] 0.05

= 0.0108 + 0.027 + 0.005

= 0.0428 or 4.28 %

Hence, probability that a randomly selected label produced by the company will contain a defective compact disk is 4.28%.

Cricetus Cricetus · Answer 2 · 2021-11-17T16:55:06+01:00

The probability of selecting defective compact disk will be "4.28%".

Given:

P(M First),

36% or 0.36

P(M Second),

54% or 0.54

P(M Third),

10% or 0.10

P(D/M First),

3% or 0.03

P(D/M Second),

5% or 0.05

P(D/M Third),

5% or 0.05

Now,

The probability,

= [tex]P(M \ I) P(D/M \ I) + P(M \ II) \ P(D/M \ II) + P(M \ III) \ P(D/M \ III)[/tex]

By substituting the values, we get

= [tex]0.36\times 0.03 + 0.54\times 0.05 + 0.10\times 0.05[/tex]

= [tex]0.0108 + 0.027 + 0.005[/tex]

= [tex]0.0428[/tex]

or,

= [tex]4.28[/tex] (%)

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