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Probability Ncert Solutions Maths class 12th Maths Class 12th Probability

45. A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, whereas the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

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alok kumar singh | Contributor-Level 10

4 months ago

45. Let, $E_{1} :$ event of time consume by $A$

$E_{2} :$ event of time consume by $B$

$E_{3} :$ event of time consume by $C$

$P (E_{1}) = 50 % = \frac{50}{100} = \frac{1}{2}$

$P (E_{2}) = 30 % = \frac{30}{100} = \frac{3}{10}$

$P (E_{3}) = 20 % = \frac{20}{100} = \frac{1}{5}$

Let, $A :$ event of producing the defective item. Therefore,

$P (A / E_{1}) = 1 % = \frac{1}{100}$

$P (A / E_{2}) = 5 % = \frac{5}{100}$

$P (A / E_{3}) = 7 % = \frac{7}{100}$

Therefore, by Baye’s theorem,

$P (E_{1} / A) =$ probability that the defective item was produced by $A$ ,

$P (E_{1} / A) = \frac{P (E_{1}) . P (A / E_{1})}{P (E_{1}) . P (A / E_{1}) + P (E_{2}) . P (A / E_{2}) + P (E_{3}) . P (A / E_{3})}$

$= \frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{2} \times \frac{1}{100} + \frac{3}{10} \times \frac{5}{100} + \frac{1}{5} \times \frac{7}{100}}$

$= \frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{100} (\frac{1}{2} + \frac{15}{10} + \frac{7}{5})} = \frac{\frac{1}{2} \times \frac{1}{100}}{\frac{1}{100} (\frac{5 + 15 + 14}{10})} = \frac{\frac{1}{2}}{\frac{34}{10}}$

$= \frac{1}{2} \times \frac{10}{34} = \frac{5}{34}$

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