View Full Version : High school math question... (im dumb)
wegs the wegs
09-29-2005, 04:15 PM
And im in college...
Machine one has an 88.38% of working. If that machine goes down machine two is the backup and also has an 88.38% chance of working. What are the odds that at least one of them is working?
LetYouDown
09-29-2005, 04:27 PM
Unless I'm misunderstanding the question...
Well, there's a matter of the amount of time...but for the sake of the question:
The probability that a machine is NOT working = 11.62%
Probability that they're both NOT working = (.1162)^2 = 1.350244%.
So: 1 - .01350244 = ~98.65%
Mr. Curious
09-29-2005, 05:09 PM
[ QUOTE ]
Unless I'm misunderstanding the question...
Well, there's a matter of the amount of time...but for the sake of the question:
The probability that a machine is NOT working = 11.62%
Probability that they're both NOT working = (.1162)^2 = 1.350244%.
So: 1 - .01350244 = ~98.65%
[/ QUOTE ]
Is this the same thing as:
88.38% machine 1 is running.
The other 11.62% of the time, there is a 88.38% that machine 2 is running.
So the equation is 88.38 + (11.62 * .8838) = 88.38 + 10.29 = 98.67?
[ QUOTE ]
Is this the same thing as:
88.38% machine 1 is running.
The other 11.62% of the time, there is a 88.38% that machine 2 is running.
So the equation is 88.38 + (11.62 * .8838) = 88.38 + 10.29 = 98.67?
[/ QUOTE ]
Yes.
1-(1-a)(1-b)=1-1+a+b-ab=a+(1-a)b
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