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Home  >>  CBSE XII  >>  Math  >>  Probability
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A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as 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 is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

$\begin{array}{1 1} 5/24 \\ 5/34 \\ 7/24 \\ 7/34 \end{array} $

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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 machine A is defective. P ($E_1$) = 1%.
Let $E_2$ be the event that the machine B is defective. P ($E_2$) = 5%.
Let $E_3$ be the event that the machine C is defective. P ($E_3$) = 7%.
Let A: be the event that an item chosen at random was an defective item.
P (that the defective item came from A) = P (A|$E_1$) = 50%.
P (that the defective item came from B) = P (A|$E_2$) = 30%.
P (that the defective item came from C) = P (A|$E_3$) = 20%.
We need to calculate that if a defective item is produced what is the probability that it came from A?
We can use Baye's theorem, according to which $P(E_1|A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)(PA|E_3)}$
Therefore P(A|$E_1$) = $\large \frac{0.05 \times 0.1}{0.05 \times 0.1 + 0.30 \times 0.05 + 0.20 \times 0.07} = \frac{0.005}{0.005+0.0150+0.014} = \frac{0.005}{0.034} = \frac{5}{34}$
answered Jun 19, 2013 by balaji.thirumalai
 

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