One glass of milk gives you 3/10 of the recommended amount of calcium per day.
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<h3>~AH1807</h3>Answer:
Probability that a randomly selected label produced by the company will contain a defective compact disk is 0.0428 or 4.28%.
Step-by-step explanation:
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
<em>Also, let D = defective compact disks</em>
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) P(D/M I) + P(M II)
P(D/M II) + P(M III)
P(D/M III)
= 0.36 0.03 + 0.54
0.05 + 0.10
0.05
= 0.0108 + 0.027 + 0.005
= 0.0428 or 4.28 %
<u><em>Hence, probability that a randomly selected label produced by the company will contain a defective compact disk is 4.28%.</em></u>