1. There are 3 machines in a factory that perform all production. Let these machines work independently of each other with probabilities 0.8, 0.9 and 0.95, respectively. a. If the random variable is defined as the number of machines running at any one time, find the probability distribution of ?. b. According to the maintenance agreement made by the factory, the repair price of a machine is fixed at 500 TL. To repair the broken machines at any time, every time the maintenance personnel are called, they maintain each of the intact ones for 50 TL. In this case, how much is this maintenance agreement expected to cost to the factory?

College Algebra
10th Edition
ISBN:9781337282291
Author:Ron Larson
Publisher:Ron Larson
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 4ECP: Show that the probability of drawing a club at random from a standard deck of 52 playing cards is...
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1. There are 3 machines in a factory that perform all production. Let these machines work independently of each other with probabilities 0.8, 0.9 and 0.95, respectively. a. If the random variable is defined as the number of machines running at any one time, find the probability distribution of ?. b. According to the maintenance agreement made by the factory, the repair price of a machine is fixed at 500 TL. To repair the broken machines at any time, every time the maintenance personnel are called, they maintain each of the intact ones for 50 TL. In this case, how much is this maintenance agreement expected to cost to the factory?

Expert Solution
Step 1

Let X = Number of machines working

Let machines be A1, A2 and A3

Let these machines work independently of each other with probabilities 0.8, 0.9 and 0.95, respectively.

 

Step 2

a) Find the probability distribution of x, if the random variable is defined as the number of machines running at any one time:

X = 0 => No machine is working and so on

X = 0 => P(X= 0) = (1-0.8)(1-0.9)(1-0.95)

                             = 0.2*0.1*0.05

                             = 0.001

X = 1 => P(X= 1) = 0.8(1-0.9)(1-0.95)+(1-0.8)(0.9)(1-0.95)+(1-0.8)(1-0.9)(0.05)

                             = 0.014

X = 2 => P(X = 2) = 0.8(0.9)(1-0.95)+(1-0.8)(0.9)(0.95)+(0.8)(1-0.9)(0.95)

                             = 0.283

X = 3 => P(X = 3) = 0.8*0.9*0.95

                            = 0.684

 

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