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Exercises · Q13

Q.A factory has three machines A, B and C producing 25%, 35% and 40% of its total output respectively. Their defective rates are 5%, 4% and 2%. An item is selected at random from the factory's output and found to be defective. Find the probability that it was produced by machine C.

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Step 1 — Prior (output-share) probabilities.

P(A)=0.25,P(B)=0.35,P(C)=0.40P(A)=0.25,\quad P(B)=0.35,\quad P(C)=0.40

Step 2 — Conditional (defective-rate) probabilities.

P(D∣A)=0.05,P(D∣B)=0.04,P(D∣C)=0.02P(D\mid A)=0.05,\quad P(D\mid B)=0.04,\quad P(D\mid C)=0.02

Step 3 — Total probability of a defective item (Law of Total Probability).

P(D)=P(A)P(D∣A)+P(B)P(D∣B)+P(C)P(D∣C)=(0.25)(0.05)+(0.35)(0.04)+(0.40)(0.02)P(D)=P(A)P(D\mid A)+P(B)P(D\mid B)+P(C)P(D\mid C)=(0.25)(0.05)+(0.35)(0.04)+(0.40)(0.02)

=0.0125+0.0140+0.0080=0.0345=0.0125+0.0140+0.0080=0.0345

Step 4 — Apply Bayes' Theorem for machine C.

P(C∣D)=P(C)P(D∣C)P(D)=0.00800.0345≈0.2319P(C\mid D)=\dfrac{P(C)P(D\mid C)}{P(D)}=\dfrac{0.0080}{0.0345}\approx0.2319 …

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