Question

Three machines in a factory produce respectively 20%, 50% and 30% of items daily. The percentage of defective items of these machines are respectively 3, 2 and 5. An item is taken at random from the production and is found to be defective, the probability that it is produced by machine A is ____.

• A. 6/31
• B. 3/31
• C. 1/31
• D. 2/31

Answer 1

The correct option is A 6/31

$\text{Probability of item produced by machine A}\text{P}\left(A\right)$
$=20$
$\text{Probability of item produced by machine B}\text{P}\left(B\right)$
$=50$
$\text{Probability of item produced by machine C}\text{P}\left(C\right)$
$=30$
$G\text{iven the item produced by A, probability of it being defective}$
$\text{P}\left(\frac{D}{A}\right)=3$
$G\text{iven the item produced by B, probability of it being defective}$
$\text{P}\left(\frac{D}{B}\right)=2$
$G\text{iven the item produced by C, probability of it being defective}$
$\text{P}\left(\frac{D}{C}\right)=5$
$P\left(\frac{A}{D}\right)=\frac{P\left(A\right)\times P\left(\frac{D}{A}\right)}{[P\left(A\right)\times P\left(\frac{D}{A}\right)]+[P\left(B\right)\times P\left(\frac{D}{B}\right)]+[P\left(C\right)\times P\left(\frac{D}{C}\right)]}$
$=\frac{0.2\times 0.03}{[0.2\times 0.03]+[0.5\times 0.02]+[0.3\times 0.05]}$
$=\frac{0.006}{0.006+0.01+0.015}$
$=\frac{0.006}{0.031}=\frac{6}{31}$