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Exercise 12.4 · Q3

Q.A firm manufactures PVC pipes in three plants, XX, YY and ZZ. The daily production volumes from the three plants XX, YY and ZZ are respectively 2000 units, 3000 units and 5000 units. It is known from past experience that 3% of the output from plant XX, 4% from plant YY and 2% from plant ZZ are defective. A pipe is selected at random from a day's total production.

(i) Find the probability that the selected pipe is defective.
(ii) If the selected pipe is defective, what is the probability that it was produced by plant YY?
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Step 1. Given. Total daily output =2000+3000+5000=10000=2000+3000+5000=10000 units, so P(X)=0.2P(X)=0.2, P(Y)=0.3P(Y)=0.3, P(Z)=0.5P(Z)=0.5. Defect rates: P(def/X)=0.03P(\text{def}/X)=0.03, P(def/Y)=0.04P(\text{def}/Y)=0.04, P(def/Z)=0.02P(\text{def}/Z)=0.02.

Step 2. Part (i) -- Total Probability. P(def)=(0.2)(0.03)+(0.3)(0.04)+(0.5)(0.02)=0.006+0.012+0.010=0.028P(\text{def})=(0.2)(0.03)+(0.3)(0.04)+(0.5)(0.02)=0.006+0.012+0.010=0.028. …

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