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12-03-2013, 10:44 AM   #1
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Probability question...desired result.

Can someone help me with a probability problem?

There are seven separate events each with a probability of a desired result. How do you figure the probability of none of these seven events having the desired result.

1) 75% of having the desired result or 3 out of 4
2) 25% or 1 out of 4
3) 25% or 1 out of 4
4) 33% or 4 out of 12
5) 36% or 4 out of 11
6) 9% or 1 out of 11
7) 8% or 1 out of 12

I'm sure that there is an excellent chance of having at least one of the seven giving the desired result; probably close to 100%. But I know it's not 100%. Yet, I can't figure out the chances of having the desired result. Or not having the desired result.

Hope I didn't confuse you. As always, many thanks.
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Quote:
 Originally Posted by BC1836 What? 2019 or 2020? Say it ain't so! You must "sneak in a day" next year. That's an order!

 12-03-2013, 11:21 AM #2 aaarcher86 DIS Veteran     Join Date: Feb 2010 Location: Columbus, OH Posts: 7,659 The answer is Skittles. __________________
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 12-03-2013, 11:29 AM #3 Patch'sD DIS Veteran   Join Date: Sep 1999 Location: Universal Only Posts: 2,206 There is 1/4 chance of the first not happening, There is 3/4 chance of the next not happening, and so on. So Combining the odds 1/4 x 3/4 x 3/4 x 8/12 x 7/11 x 10/11 x 11/12 Multiply the numerators and the denominators and we get 55,540 (Numerator) of the desired result not happening out of 1,115,136 Chances. So the odds of not getting the result is .049 (4.9%) or roughly 5/100 or 1/20. So the chance of none of the outcomes reaching the desired result in 5%
 12-03-2013, 12:33 PM #4 GRUMPY PIRATE First rule, always!!     Join Date: Aug 2007 Location: Earth, north America, left coast, San Diego Posts: 10,536 There is a 100% probability that you are getting the DIS to do your homework! __________________ What?
12-03-2013, 01:19 PM   #5
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Quote:
 Originally Posted by Patch'sD Multiply the numerators and the denominators and we get 55,540 (Numerator) of the desired result not happening out of 1,115,136 Chances. So the odds of not getting the result is .049 (4.9%) or roughly 5/100 or 1/20. So the chance of none of the outcomes reaching the desired result in 5%
This.
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12-03-2013, 01:24 PM   #6
KennesawNemo
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Quote:
 Originally Posted by ladybebop can someone help me with a probability problem? There are seven separate events each with a probability of a desired result. How do you figure the probability of none of these seven events having the desired result. 1) 75% of having the desired result or 3 out of 4 2) 25% or 1 out of 4 3) 25% or 1 out of 4 4) 33% or 4 out of 12 5) 36% or 4 out of 11 6) 9% or 1 out of 11 7) 8% or 1 out of 12 i'm sure that there is an excellent chance of having at least one of the seven giving the desired result; probably close to 100%. But i know it's not 100%. Yet, i can't figure out the chances of having the desired result. Or not having the desired result. Hope i didn't confuse you. As always, many thanks.
=0.25*0.75*0.75*0.67*0.64*0.91*0.92
=0.05048316
=5.05%

12-03-2013, 01:36 PM   #7
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Quote:
 Originally Posted by GRUMPY PIRATE There is a 100% probability that you are getting the DIS to do your homework!
There is a 100% probability that this 54 year old woman is not in school. (Yes, I know some 50+ year olds are in school. I'm not one of them.)

Thanks everyone. I wanted to multiply the 75% by 25% etc. But I knew the answer had to be more than 75%. So mulitplying the 25% et al makes more sense.

BTW, this was a real life occurence regarding the NFL. I knew the chances of this happening were very slight, however it happened last year.
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Next trip: When my next grandchild turns five. Probably around the turn of the next decade.

Quote:
 Originally Posted by BC1836 What? 2019 or 2020? Say it ain't so! You must "sneak in a day" next year. That's an order!

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