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What is probability that u/v/w and x/y/z are reciprocal [#permalink]
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30 Sep 2008, 00:02
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What is probability that u/v/w and x/y/z are reciprocal fractions? 1: v,w,y,z are each randomly chosen from the first 100 positive integers. 2: The product of U X is the median of 100 consecutive integers.
Hint: we need to prove that ux = vwyz



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Re: reciprocal probability [#permalink]
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30 Sep 2008, 02:36
not sure whether my answer is right.
From stmt2 UX = 50.5
If, I combing stmt1 with stmt2, it simply says that UX cannot be 50.5 because U and X are integers. Hence, the required probability will be 0.
Hence, C.



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Re: reciprocal probability [#permalink]
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30 Sep 2008, 12:48
yes, it should be C.
I  v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient
II  ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z
I and II  we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.
thus C.



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Re: reciprocal probability [#permalink]
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30 Sep 2008, 20:05
aim2010 wrote: yes, it should be C.
I  v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient
II  ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z
I and II  we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.
thus C.
The product of U X is the median of 100 consecutive integers. How can the median of 100 consecutive integers take the form of x.5? median of consecutive integers is an integer. correct? Still that does not tell us any thing about product vwyz



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Re: reciprocal probability [#permalink]
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30 Sep 2008, 20:20
icandy wrote: aim2010 wrote: yes, it should be C.
I  v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient
II  ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z
I and II  we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.
thus C.
The product of U X is the median of 100 consecutive integers. How can the median of 100 consecutive integers take the form of x.5? median of consecutive integers is an integer. correct? Still that does not tell us any thing about product vwyz median of an odd number of consecutive integers is an integer. for even number of consecutive integers median is the mean of the middle 2 which would give you a fraction. like for { 1, 2, 3, 4 } median is 2.5



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Re: reciprocal probability [#permalink]
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30 Sep 2008, 20:24
aim2010 wrote: icandy wrote: aim2010 wrote: yes, it should be C.
I  v w y z could be any of the first 100 positive integers. no information about u and x thus insufficient
II  ux could be any number of the form x.5 where x is any integer. for example, 0.5, 1.5, 2.5, 3.5 etc.. like if we take "first" 100 consecutive integers, ux is 50.5. doesn't tell anything about v w y z
I and II  we have to find out probability of us being equal to vwyz; product of v w y z (all integers) can never be a fraction of the form x.5. probability is 0.
thus C.
The product of U X is the median of 100 consecutive integers. How can the median of 100 consecutive integers take the form of x.5? median of consecutive integers is an integer. correct? Still that does not tell us any thing about product vwyz median of an odd number of consecutive integers is an integer. for even number of consecutive integers median is the mean of the middle 2 which would give you a fraction. like for { 1, 2, 3, 4 } median is 2.5 Shit! I need some caffeine. Thanks for making it obvious




Re: reciprocal probability
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30 Sep 2008, 20:24






