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

42. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B?

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alok kumar singh | Contributor-Level 10

4 months ago

42. Let, $E_{1} :$ event items produced by $A$

$E_{2} :$ event items produced by $B$

$X :$ event that produced item was found to be defective

$P (E_{1}) = 60 % = \frac{3}{5}$

$P (E_{2}) = 40 % = \frac{2}{5}$

$P (X / E_{1}) = P$ (items produced by machine $A$ which is defective) $= 2 % = \frac{2}{100}$

$P (X / E_{2}) = P$ ( items produced by machine $B$ which is defective) $= 1 % = \frac{1}{100}$

Therefore, by Baye’s theorem,

$P (E_{2} / X) =$ probability that the randomly selected item was from machine $B$ ,which is defective,

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

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

$= \frac{\frac{2}{500}}{\frac{6}{500} + \frac{2}{500}}$

$= \frac{2}{500} \times \frac{500}{8} = \frac{1}{4}$

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