Answer to Question #301736 in Statistics and Probability for jitendra

Question

A manufacturing firm produces steel pipes in three plants with daily

production volumes of 500, 1000 and 2000 units respectively.

According to past experience, it is known that the fraction of defectives output produced by the three plants are respectively 0.005,

0.008 and 0.010. If a pipe is selected from a day’s total production

and found to be defective, find out (a) from which plant the pipe

comes (b) what is the probability that it comes from first plant?

Answer 1

(a)

By summing the production volumes from the three plants,;

Total production output = 500+1000+2000 = 3500

In order to determine which plant the pipe came from, we compute the probabilities of production. In this case, the probability that the defective pipe selected comes from:

P(A) = 500/3500 = 1/7

P(B) = 1000/3500 = 2/7

P(C) = 2000/3500 =4/7

(b)

Given the probability of defective pipes by:

Plant A = (A|E) = 0.005

Plant B = (B|E) = 0.008

Plant C = (C|E) = 0.010

Also, from (a), we obtained the following probabilities of production:

P(A) = 500/3500 = 1/7

P(B) = 1000/3500 = 2/7

P(C) = 2000/3500 =4/7

Now, we need to find the probability that the defective pipe comes from first plant.

By using the Bayes theorem, we know that:

P(E/A)= (P(A).P(A|E))/(P(A).P(A|E)+P(B).P(B|E)+P(C).P(C|E))

= 1/7*(0.005)/(1/7x0.005 + 2/7x0.008 + 4/7x0.010)

= 5/61

Therefore, the probability that it comes from first plant = 5/61