Question

A manufacturer has three machine operators A, B an C. The first operatot A produces 1% defective items, where as the other two operators B anc C produce 5% of the time, B on the job for 30% of the time and C on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?

Answer 1

Let $E_1$ : the event that item is produced by machine A

$E_2$ : the event that item is produced by machine B

and $E_3$ : the event that item is produced by machine C

Here, $E_1,E_2,and{E}_3$ are mutually exclusive and exhaustive events.

Moreover,

$P\left(E_1\right)=50\%=\frac{50}{100},P\left(E_2\right)=30\%=\frac{30}{100}andP\left(E_3\right)=20\%=\frac{20}{100}$
Let E : The event that item chosen is found to be defective',

$\therefore$ $P\left(\frac{E}{E_1}\right)=\frac{1}{100},P\left(\frac{E}{E_2}\right)=\frac{5}{100},andP\left(\frac{E}{E_3}\right)=\frac{7}{100}.$

By using Baye's therorem, we obtain

$P\left(\frac{E_1}{E}\right)=\frac{P\left(\frac{E}{E_1}\right)P\left(E_1\right)}{P\left(\frac{E}{E_1}\right)P\left(E_1\right)+P\left(\frac{E}{E_2}\right)P\left(E_2\right)+P\left(\frac{E}{E_3}\right)P\left(E_3\right)}=\frac{\frac{1}{100}\times \frac{50}{100}}{\frac{1}{100}\times \frac{50}{100}+\frac{5}{100}\times \frac{30}{100}+\frac{7}{100}\times \frac{20}{100}}=\frac{50}{50+150+140}=\frac{50}{340}=\frac{5}{34}$