A company produces mixed types of widgets.? An inspector randomly select 3 widgets from a production line assumed to

An inspector randomly select 3 widgets from a production line assumed to relinquish 90 % satisfactory and 10 % untisatisfactoy output. You may assume Successive power events to be independent.
A. Construct a probability tree diagram for the experience identity all the space, compute the amalgamated 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 smallest 1, 6. at most 2

Answers: Dear jay r,

A. You should label the branches of the tree beside either and S (for satisfactory) or U (for unsatisfactory), along near the probability for that branch given the results of earlier branches (you don't in actual fact have to verbs about the nearer results in this problem because you are told that part events are independent of each other). At the ends of the secure of branches you should also put the respective end results next to the corresponding joint probabilities (e.g., 2 S, 1 U, 0.081). Here are the label, but you'll have to make a payment 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 make the addition of the probabilities there for the results that you want (e.g., adjectives branches ending beside 1 U would be the cases with exactly one unsatisfactory widget).

1. P(0 U) = 0.729 .

2. I guess you intended "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 .

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