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zwackerm
Hold the door!
Joined: Sun Jun 01, 2014 10:26 pm
Posts: 21462
Location: West Chester, Pennsylvania
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 Help with a math problem? If the odds of event A happening are 98/100, the odds of event B happening are 999/1000, and the odds of event C happening are 21/28, what are the odds of none of the events happening?
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| Thu Mar 16, 2017 10:32 am |
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Alex Y.
Top Poster
Joined: Fri Oct 15, 2004 4:47 pm
Posts: 5812
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Re: Help with a math problem? (1-(98/100))*(1-(999/1000)*(1-(21/28))
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| Thu Mar 16, 2017 11:05 am |
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zwackerm
Hold the door!
Joined: Sun Jun 01, 2014 10:26 pm
Posts: 21462
Location: West Chester, Pennsylvania
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Re: Help with a math problem? That would be .24. Definitely wrong.
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| Thu Mar 16, 2017 11:53 am |
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Shack
Devil's Advocate
Joined: Sun Jul 31, 2005 2:30 am
Posts: 40245
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Re: Help with a math problem? Instead of multiplying the chance of each happening you multiply the chance of each not happening
Event A = 2/100 or .02
Event B = 1/1000 or .001
Event C = 7/28 or .25
.02 * .001 * .25 = .000005
_________________Shackâs top 50 tv shows - viewtopic.php?f=8&t=90227
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| Thu Mar 16, 2017 12:15 pm |
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_axiom
The Wall
Joined: Wed Jan 26, 2005 10:50 am
Posts: 16163
Location: Croatia
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Re: Help with a math problem? Shack wrote: Instead of multiplying the chance of each happening you multiply the chance of each not happening
Event A = 2/100 or .02
Event B = 1/1000 or .001
Event C = 7/28 or .25
.02 * .001 * .25 = .000005 That's what Alex wrote. zwackerm just didn't put it in the calculator as he should've.
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| Thu Mar 16, 2017 12:28 pm |
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zwackerm
Hold the door!
Joined: Sun Jun 01, 2014 10:26 pm
Posts: 21462
Location: West Chester, Pennsylvania
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Re: Help with a math problem? Shack wrote: Instead of multiplying the chance of each happening you multiply the chance of each not happening
Event A = 2/100 or .02
Event B = 1/1000 or .001
Event C = 7/28 or .25
.02 * .001 * .25 = .000005 Thanks Shack and Alex! Sorry I misread your answer.
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| Thu Mar 16, 2017 1:18 pm |
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