(1 point) A tent manufacturer knows from past data that there is an 12.9% chance
ID: 3040578 • Letter: #
Question
(1 point) A tent manufacturer knows from past data that there is an 12.9% chance of having a defect in the zipper of a tent and a 3.2% chance of having a defect in the seam sealing treatment on the tent. The machines that do these parts of the tent manufacturing are independent of each other, but every tent gets both a zipper and a seam sealing treatment (a) What is the probability that the tent will have a defective zipper AND a defective seam sealing treatment? 0.004128 (b) What is the probability that the tent will have neither a defective zipper nor a defective seam sealing treatment? (c) What is the probability of having at least one of the two possible defects? Hint: It might be easier to use the complement rule on this one: P(notA) 1-PA) (d) What is the probability of getting exactly 1 of the defects? Hint: If you have two events, A and B, and you want exactly one then calculate the probability of getting A and not B OR the probability of getting B and not A. (e) What is the probability of getting a zipper defect but not a seam sealing defect? (1 What is the probability of getting a seam sealing defect but not a zipper defect?Explanation / Answer
Answer to the question is as follows:
a.
p(defect zipper) = .129
p(defect seam) = .032
The 2 defects are caused independent of each other. So, P( A and B) = P(A)*P(B)
a. P( defect zipper and seam) = P(defect zipper)*P(defect seam) = .129*.032 = .000428
b. P(neither of defects) = P(A' and B') = (1-.129)*(1-.032) = .8431
c. P(atleast one of the two defect) = 1 - P( neither of defects) = 1-.8431 = .1569
d. P(exactly 1 defect) = P(defect seam)*P(no defect zipper) + P(no defect seam)*P(defect zipper) = (1-.129)*(.032) + (.129)*(1-.032) = .1527
e. P( defect zipper but not a seam defect) = .129*(1-.032) = .1249
f.P( defect seam but not a seam zipper) = (1-.129)*(.032) = .0279
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