Please help me to understand this example and why he choose plant 2 given the standard ExampleA certain item is manufactured at two plants. Plant 1 makes 70% ofthe requirement and Plant 2 makes 30%. From Plant 1, 90% meet therequired standard and from Plant 2 only 80%.What is the probability that an item purchased by a customer meetsthe standard?Consider, st is the event that an item is up to standardPl is the event that the item is made in Plant 1P2 is the event that the item is made in Plant 2P(P1) = 0.7P(P2) = 0.3P(st P1) = 0.9P(st P2) = 0.8P(st) = P(st | P1). P(P1) + P(st | P2). P(P2)= 0.9 x 0.7 + 0.8 x 0,3 = 0.63 + 0.24 = 0.87Reliability Evaluation of Engineering Systems32Tuesday, 08 June 2021ENG Example (cont.)Probability that the item was made in Plant 2,GIVEN that is standard,P(P2nst)P(st P2). P(P2)0.8 x 0.3P(P2 st) =P(st)P(st)0.87= 0.276P1P20.70.3stststst0.90.10.80.20.070.240.060.63Reliability Evaluation of Engineering Systems33Tuesday, 08 June 2021ENG

This is an example solved by using the BAYES' theorem. In the first part of the example, the…

Probability that the item was made in Plant 2, GIVEN that is standard, P(P2nst) P(st P2). P(P2) 0.8 x 0.3 P(P2 st) = %3D P(st) P(st) 0.87 = 0.276 %3D P1 P2 0.7 0.3 st st st st 0.9 0.1 0.8 0.2 0.07 0.24 0.06 0.63 Tuuluation of Engineering System:

Probability that the item was made in Plant 2, GIVEN that is standard, P(P2nst) P(st P2). P(P2) 0.8 x 0.3 P(P2 st) = %3D P(st) P(st) 0.87 = 0.276 %3D P1 P2 0.7 0.3 st st st st 0.9 0.1 0.8 0.2 0.07 0.24 0.06 0.63 Tuuluation of Engineering System:

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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