Probability and chance of failure

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Homework Help Overview

The discussion revolves around a probability problem related to an experiment with a 96% success rate. The original poster is interested in determining the chance of experiencing at least one failure when the experiment is conducted 100 times.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the probability of failure by considering both "exactly 1 failure" and "at least 1 failure" scenarios. There is a suggestion to approach the problem by calculating the opposite scenario first.

Discussion Status

Some participants have provided calculations related to the binomial distribution, indicating a productive direction in the discussion. However, there is no explicit consensus on the final approach or interpretation of the problem.

Contextual Notes

The original poster expresses fatigue, which may affect their reasoning. There is a mention of not wanting to crowd the general math forum, indicating an awareness of the forum's structure.

Zach_C
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Ok, not really homework but I did not want to crowd general math.

I have an experiment that has 96% chance of working. If I try it one hundred times what is the chance of one being a failure.

I know with a one in two shot it is .5^2=.25 You now have a 25 percent chance of failure. Any help. Sorry at the moment my brain seems dead.
 
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Oh my God I'm an idiot. I really am more tired than this thread let's on. :zzz:
 
Try solving the opposite problem.
 
Zach_C said:
Ok, not really homework but I did not want to crowd general math.

I have an experiment that has 96% chance of working. If I try it one hundred times what is the chance of one being a failure.

I know with a one in two shot it is .5^2=.25 You now have a 25 percent chance of failure. Any help. Sorry at the moment my brain seems dead.
Do you want the probability for "EXACTLY 1" Failure or for "AT LEAST 1" Failure?? Both can be determined from the Binomial Distribution:
Prob{Exactly 1 Failure} = (99)*{(0.96)^(99)}*{(1 - 0.96)^(1)} =
= (0.0696) = (6.96 %)
Prob{At Least 1 Failure} = 1 - Prob{Exactly 0 Failures} = 1 - (0.96)^100 =
= (0.983) = (98.3 %)


~~
 
Last edited:

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