Q. 75.0( 2 Votes )

A factory has two machines A and B. Past records show that the machine A produced 60% of the items of output and machine B produce 40% of the items. Further 2% of the items produced by machine A were defective and 1% produced by machine B were defective. If an item is drawn at random, what is the probability that it is defective?

Answer :

Given:


Machine A produced 60% of the total items.


Machine B produced 40% of the total items.


2% of the items produced by machine A were defective.


1% of the items produced by machine B were defective.


There are two mutually exclusive ways to draw a defective item produced by one of the two machines –


a. Item was produced by machine A, and then, the item is defective


b. Item was produced by machine B, and then, the item is defective


Let E1 be the event that the item was produced by machine A and E2 be the event that the item was produced by machine B.


As 60% of the total items were produced by machine A, we have




Similarly, as 40% of the total items were produced by machine B, we have




Let E3 denote the event that the item is defective.


Hence, we have



2% of the items produced by machine A are defective.



We also have



1% of the items produced by machine A are defective.



Using the theorem of total probability, we get


P(E3) = P(E1)P(E3|E1) + P(E2)P(E3|E2)






Thus, the probability of the item being defective is.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :