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Home  >>  CBSE XII  >>  Math  >>  Probability
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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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  • Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability $P(E_i|A)$ for any event A associated w/ $E_i$ using the Bayes theorem as follows: \(\;P(E_i/A)\)=\(\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}\)
Let $E_1$ be the event that the item is produced by machine A. P($E_1$) = 60% = $\large\frac{60}{100} = \frac{3}{5}$
Let $E_2$ be the event that the item is produced by machine A. P($E_2$) = 40% = $\large\frac{40}{100} = \frac{2}{5}$
Let A be the even that we choose a defective item at random:
P (the defective item came from machine A) = P (A|$E_1$) = 2% = $\large\frac{2}{100}$
P (the defective item came from machine B) = P (A|$E_2$) = 1% = $\large\frac{1}{100}$
We need to find the probability that a defective item randomly selected was produced by machine B
We can use Baye's theorem, according to which $P(E_2|A) = \large\frac{P(E_2)(P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)}$
P ($E_2$|A) = $\Large\frac{\frac{1}{100}.\frac{3}{5}}{\frac{1}{100}.\frac{3}{5} + \frac{2}{100}.\frac{2}{5}} $ = $\Large \frac{\frac{2}{500}}{\frac{6}{500}+\frac{2}{500}} $=$\large\frac{2}{8} = \frac{1}{4}$
answered Jun 19, 2013 by balaji.thirumalai
 

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