High school math question... (im dumb)

wegs the wegs · #1 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 · #2 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 · #3 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?

#4 09-29-2005, 05:44 PM

[ 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