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A company manufactured a product through 4 units. A, B, C and D. The probability that a product is manufactured are 310,15,110\frac{3}{10},\frac{1}{5},\frac{1}{10} and 25\frac{2}{5} by units A, B, C and D respectively. The probability that it will be defective are 14,13\frac{1}{4},\frac{1}{3} and 112\frac{1}{12} if it is produced by A, B, C respectively and there is no defective from unit D. A product is chosen, it is found to be defective.

Match the LIST-I with LIST-II

LIST-ILIST-II
A. The Probability that the defective product is from AI. 4/9
B. The Probability that the defective product is from BII. 0
C. Probability of defective productIII. 1/2
D. Probability that the defective product is from DIV. 3/20

Choose the correct answer from the options given below:

Solution

✅ Correct Option: 2

P(defective) =310⋅14+15⋅13+110⋅112+0=9+8+1120=320=\frac{3}{10}\cdot\frac14+\frac15\cdot\frac13+\frac1{10}\cdot\frac1{12}+0=\frac{9+8+1}{120}=\frac{3}{20}. By Bayes' theorem, P(A | defective) =9/12018/120=12=\frac{9/120}{18/120}=\frac12, P(B | defective) =818=49=\frac{8}{18}=\frac49, and P(D | defective) =0=0.

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