A factory has three machines, A, B, and C, for producing items. Machine A produces 50%, B produces 30%, and C produces 20%. If a randomly selected item from the factory's output is found to be defective, what is the probability that B made it?

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A factory has three machines, A, B, and C, for producing items. Machine A produces 50%, B produces 30%, and C produces 20%. If a randomly selected item from the factory's output is found to be defective, what is the probability that B made it?

It is also known that 3% of the items produced by machine A are defective, as are 2% of the items produced by machine B and 1% of the items from machine C.

If a randomly selected item from the factory's output is found to be defective, what is the probability that it was produced by machine B? Also, if a randomly selected chip is found not to be defective, what is the probability that it came from machine B?

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Answer 1 · 2018-02-11T06:28:43

Given the item is defective, probability of it being produced by B $= 0.26$ $=0.26$

Explanation:

Prior information

Probability of an item produced by Machine A $P (A) = 50 % = 0.5$ $P(A)=50%=0.5$
Probability of an item produced by Machine B $P (B) = 30 % = 0.3$ $P(B)=30%=0.3$
Probability of an item produced by Machine C $P (C) = 20 % = 0.2$ $P(C)=20%=0.2$

On the basis of additional information
Given the item produced by A, probability of it being defective $P (\frac{D}{A}) = 3 % = 0.03$ $P(D/A)=3% = 0.03$
Given the item produced by B, probability of it being defective $P (\frac{D}{B}) = 2 % = 0.02$ $P(D/B)=2% = 0.02$
Given the item produced by C, probability of it being defective $P (\frac{D}{C}) = 1 % = 0.01$ $P(D/C)=1% = 0.01$

Given the item is defective, probability of it being produced by B

$P (\frac{B}{D}) = \frac{P (B) \times P (\frac{D}{B})}{[P (B) \times P (\frac{D}{B})] + [P (A) \times P (\frac{D}{A})] + [P (C) \times P (\frac{D}{C})]}$ $P(B/D)=[P(B)xxP(D/B)] /[[P(B) xx P(D/B)]+[P(A) xx P(D/A)]+[P(C) xx P(D/C)]$

$= \frac{0.3 \times 0.02}{[0.3 \times 0.02] + [0.5 \times 0.03] + [0.2 \times 0.01]}$ $=[0.3xx0.02]/[[0.3xx0.02]+[0.5xx0.03]+[0.2xx0.01]$
$\frac{0.006}{0.006 + 0.015 + 0.002} = \frac{0.006}{0.023} = 0.26$ $0.006/[0.006+0.015+0.002]=0.006/0.023=0.26$

Given the item is defective, probability of it being produced by B $= 0.26$ $=0.26$

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A factory has three machines, A, B, and C, for producing items. Machine A produces 50%, B produces 30%, and C produces 20%. If a randomly selected item from the factory's output is found to be defective, what is the probability that B made it?

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Explanation:

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