What is the Success Rate of an Overloaded Industrial Hoist?

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Homework Help Overview

The discussion revolves around the probability of success for an overloaded industrial hoist, specifically focusing on the failure rates of its upper and lower attachment hooks under excessive weight conditions. The subject area pertains to probability theory and risk assessment in mechanical systems.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are examining the independence of the failure probabilities of the hoist's components and discussing how to calculate the overall probability of success based on these assumptions. There are questions about whether the components can be treated as independent events.

Discussion Status

The discussion is ongoing, with participants providing insights into the independence of the probabilities and the implications for calculating the overall success rate. Some participants suggest that the probabilities can be assumed independent unless stated otherwise, while others express uncertainty about this assumption.

Contextual Notes

There is a lack of explicit information regarding the dependency of the failure probabilities, which is central to the calculations being discussed. Participants are navigating this uncertainty as they explore the problem.

korr2221
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An industrial hoist is being used in an emergency job where the weight exceeds the design limits of two of its components. For the amount of weight being lifted, the probability that the upper attachment hook will fail is 0.20. The probability that the lower hook will fail is 0.10. What is the probability that the hoisting job will be successfully complete?

no equations

P(a') = 0.20
P(b') = 0.10
P(a)= 1-0.20 = .80
P(b)= 1-.0.10 = .90

P(a+b)= 0.8 + 0.9 - 0.8*0.9 = 1.7- 0.72 = .98

someone suggested that this problem is indep.
if this is indep

then P(ab) = 0.8*0.9 = 0.72...

can someone review over this and check?
 
Last edited:
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hi korr2221! :smile:

if they're independent, P(ab) = P(a)P(b) :wink:
 


tiny-tim said:
hi korr2221! :smile:

if they're independent, P(ab) = P(a)P(b) :wink:

from what I've read it doesn't seem independent...?
 
If they're not independent, the question would have to tell you how they're dependent, and it doesn't.

I think you can safely assume that probabilities are independent unless either the question says they're not, or it's obvious that they're not.
 

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