A company manufactured a product through 4 units. A, B, C and D. The probability that a product is manufactured are \( \displaystyle\frac{3}{10},\displaystyle\frac{1}{5},\displaystyle\frac{1}{10} \) and \( \displaystyle\frac{2}{5} \) by units A, B, C and D respectively. The probability that it will be defective are \( \displaystyle\frac{1}{4},\displaystyle\frac{1}{3} \) and \( \displaystyle\frac{1}{12} \) 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-I | LIST-II | ||
| A. | The Probability that the defective product is from A | I. | 4/9 |
| B. | The Probability that the defective product is from B | II. | 0 |
| C. | Probability of defective product | III. | 1/2 |
| D. | Probability that the defective product is from D | IV. | 3/20 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
Let E be the event that the product is defective. The probability of each unit producing a defective product is:
\( \displaystyle P(A)P(E|A)=\frac{3}{10}\cdot\frac{1}{4}=\frac{9}{120} \)
\( \displaystyle P(B)P(E|B)=\frac{1}{5}\cdot\frac{1}{3}=\frac{8}{120} \)
\( \displaystyle P(C)P(E|C)=\frac{1}{10}\cdot\frac{1}{12}=\frac{1}{120} \)
\( \displaystyle P(D)P(E|D)=\frac{2}{5}\cdot 0=0 \)
C. Probability of defective product
This matches IV. 3/20.
By the theorem of total probability:
\( \displaystyle P(E)=\frac{9+8+1+0}{120} \)
\( \displaystyle =\frac{18}{120} \)
\( \displaystyle =\frac{3}{20} \)
A. The Probability that the defective product is from A
This matches III. 1/2.
By Bayes' theorem:
\( \displaystyle P(A|E)=\frac{9/120}{18/120} \)
\( \displaystyle =\frac{1}{2} \)
B. The Probability that the defective product is from B
This matches I. 4/9.
\( \displaystyle P(B|E)=\frac{8/120}{18/120} \)
\( \displaystyle =\frac{4}{9} \)
D. Probability that the defective product is from D
This matches II. 0.
Unit D produces no defective products, so P(D|E) = 0.
So the matching is A-III, B-I, C-IV, D-II.
Shortcut: Using the ratio method, the shares of the defectives are in the ratio 9 : 8 : 1 : 0 (out of 18), so P(A|E) = 9/18 = 1/2 and P(B|E) = 8/18 = 4/9 without writing Bayes' formula.