Answer to Question #202269 in Statistics and Probability for RAGHAV SOOD

Question

A factory produces steel pipes in three plant with daily production volumes of 500, 

1000 and 2000 units respectively from each of the plants. From the past experience it 

is known that the fraction of defective outputs produced by three plants are 

respectively 0.005, 0.008 and 0.010. If a pipe is selected at random from a day’s total production and founded to be defective, from which plant is that likely to 

have came?

Answer 1

actual Question is

A manufacturing firm produces steel pipes in three plants with daily production volume of 500,1000 and 2000 units,respectively.According to past experience,it is known that the fraction of defective outputs produced by the three plants are 0.005,0.008 and 0.010,respectively.If a pipe is selected from a day's total production and found to be defective,

then

(i)find the probability that it came from the first plant,

(ii)identity most likely plant which has produced it.

Solution

Assume that a manufacturing firm produces steel pipes in three plants A, B and C with daily production volume of 500,1000 and 2000 units,respectively.

We have probability of production

$P(A)= \frac{500}{500+1000+2000}=\frac{1}{7}\\$

$P(A)= \frac{1000}{500+1000+2000}=\frac{2}{7}\\$

$P(A)= \frac{2000}{500+1000+2000}=\frac{4}{7}$

Probability of defective pipes

$P(A|E)=0.005\\ P(B|E)=0.008\\ P(C|E)=0.010\\$

(i) By Bayes’ Theorem, probability of defective pipe from the first plant

$P(E|A)=\frac{P(A)P(A|E)}{P(A)P(A|E)+P(B)P(B|E)+P(C)P(C|E)}\\\\\\\ .\\ P(E|A)=\frac{\frac{1}{7}(0.005)}{\frac{1}{7}(0.005)+\frac{2}{7}(0.008)+\frac{4}{7}(0.010)}\\=\frac{5}{61}$

(ii)Probability of defective pipe from the second plant

$P(E|B)=\frac{P(B)P(B|E)}{P(A)P(A|E)+P(B)P(B|E)+P(C)P(C|E)}\\\\\\\ .\\ P(E|B)=\frac{\frac{2}{7}(0.008)}{\frac{1}{7}(0.005)+\frac{2}{7}(0.008)+\frac{4}{7}(0.010)}\\=\frac{16}{61}$

Probability of defective pipe from the third plant

$P(E|C)=\frac{P(C)P(C|E)}{P(A)P(A|E)+P(B)P(B|E)+P(C)P(C|E)}\\\\\\\ .\\ P(E|C)=\frac{\frac{4}{7}(0.010)}{\frac{1}{7}(0.005)+\frac{2}{7}(0.008)+\frac{4}{7}(0.010)}\\=\frac{40}{61}$

The defective pipe comes most likely from the third plant.