A company produces various types of widgets.?

An inspector randomly selects 3 widgets from a production line assumed to yield 90 % satisfactory and 10 % untisatisfactoy output. You may assume Successive quality events to be independent.


A. Construct a probability tree diagram for the experience identity all the space, compute the joint probability for each event.





B. Find the probabilities for the following number unsatisfactory items:


1- none , 2.exactly ,3. exactly 2, 4. exactly 3, 5. at least 1, 6. at most 2

A company produces various types of widgets.?
Dear jay r,





A. You should label the branches of the tree with either and S (for satisfactory) or U (for unsatisfactory), along with the probability for that branch given the results of earlier branches (you don't actually have to worry about the earlier results in this problem because you are told that quality events are independent of each other). At the ends of the chain of branches you should also put the respective end results with the corresponding joint probabilities (e.g., 2 S, 1 U, 0.081). Here are the labels, but you'll have to add a better tree drawing if you need it.





S, 0.9


__ S, 0.9


____ S, 0.9 ::: 3 S, 0 U, 0.729


____ U, 0.1 ::: 2 S, 1 U, 0.081


__ U, 0.1


____ S, 0.9 ::: 2 S, 1 U, 0.081


____ U, 0.1 ::: 1 S, 2 U, 0.009


U, 0.1


__ S, 0.9


____ S, 0.9 ::: 2 S, 1 U, 0.081


____ U, 0.1 ::: 1 S, 2 U, 0.009


__ U, 0.1


____ S, 0.9 ::: 1 S, 2 U, 0.009


____ U, 0.1 ::: 0 S, 3 U, 0.001








B. Look at the ends of the branches and add the probabilities there for the results that you want (e.g., all branches ending with 1 U would be the cases with exactly one unsatisfactory widget).





1. P(0 U) = 0.729 .





2. I guess you meant "exactly 1" unsatisfactory item here. If so,





P(1 U) = 0.081 + 0.081 + 0.081 = 0.243 .





3. P(2 U) = 0.009 + 0.009 + 0.009 = 0.027 .





4. P(3 U) = 0.001





5. P(at least 1 U) = P(1 U) + P(2 U) + P(3 U)


= 0.243 + 0.027 + 0.001


= 0.271 .





This could also be found as 1 - P(0 U) = 1 - 0.729 = 0.271 .





6. P(at most 2 U) = P(0 U) + P(1 U) + P(2 U) = 1 - P(3 U)


= 1 - 0.001


= 0.999 .








P. S. If you post such questions in the mathematics category, you'll have more people try to help you, but in the category you asked this question, most people will just ignore you.