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# what is the probab of selecting a " clean" number from a set

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Manager
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what is the probab of selecting a " clean" number from a set [#permalink]

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18 Feb 2005, 09:00
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what is the probab of selecting a " clean" number from a set of integers containing all multiples of 3 between 1 and 99, inclusive?

(1) A " clean" no is an integer divisible only by 2 factors, one of which is greater than 2

(2) A " clean" no must be odd

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Manager
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18 Feb 2005, 09:05
I am gonna go off on a limb and chose A
because all multiples of 3 are not odd 3,6,9,12
and 3*1 works; 2*3 will also work....terefore negating statement two
(A)

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Manager
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18 Feb 2005, 09:14
Could you elaborate why stmt 2 is insuff.....I am lost.....

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Manager
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18 Feb 2005, 09:46
Statement 2 is in sufficient cause it would mean that the clean numbers are 3, 5,7...
statement 1 shows that it is 3,5, 6,7..
does that show why there is at least 1 non odd number?

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Senior Manager
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18 Feb 2005, 19:48

The question is:
what is the probab of selecting a " clean" number from a set of integers containing all multiples of 3 between 1 and 99, inclusive?

So in my opinion all you need is a definition of what a clean number is. From both statements you can get a definite number of integers that fulfill the criterion mentioned...so therfore I think the answer is D.

And statement 2 should be sufficient because all you have to do is pick the odd number which are multiples of 3 and divide them by all the odd numbers between 1 and 99...don't see why that isn't sufficient
_________________

"No! Try not. Do. Or do not. There is no try.

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Manager
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18 Feb 2005, 20:20
greenandwise wrote:

The question is:
what is the probab of selecting a " clean" number from a set of integers containing all multiples of 3 between 1 and 99, inclusive?

So in my opinion all you need is a definition of what a clean number is. From both statements you can get a definite number of integers that fulfill the criterion mentioned...so therfore I think the answer is D.

And statement 2 should be sufficient because all you have to do is pick the odd number which are multiples of 3 and divide them by all the odd numbers between 1 and 99...don't see why that isn't sufficient

definitely, i agree with u my man...the basic object is to find 'probability' not 'validity'

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Intern
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19 Feb 2005, 10:20
This question is similar to another one posted here somewhere. Statement 1 gives a complete (in itself) definition of a clean number whereas Statement 2 only describes what could be one of many attributes of a clean no.

A it is!

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Director
Joined: 19 Nov 2004
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19 Feb 2005, 12:24
The set is clearly defined. Whatever the restrictions the statements 1 and 2 impose, we can still find a single probability. So it is D

Note: If the set were infinite, we can't find a probability.

EDIT:
On rethink - A is probably it.
Statement 2 does not clearly define a Clean number.
A clean number must be odd does not imply all odd numbers are clean numbers.

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VP
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19 Feb 2005, 22:52
swath20 wrote:
what is the probab of selecting a " clean" number from a set of integers containing all multiples of 3 between 1 and 99, inclusive?

(1) A " clean" no is an integer divisible only by 2 factors, one of which is greater than 2
(2) A " clean" no must be odd

using each statement seperately, both are suff to ans independently. but if you, take stat 1 into consideration 2 seems incomplete. we know that a clean number is only 3 and each st must give a specific answer. therefore, OA should be A. however, we cannot totaly disagree with the other part of the discussions.

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Senior Manager
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20 Feb 2005, 04:16
MA wrote:
thearch wrote:
IMO, That's right!

IMO=?

In my opinion

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Senior Manager
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23 Feb 2005, 09:13
swath20 wrote:
what is the probab of selecting a " clean" number from a set of integers containing all multiples of 3 between 1 and 99, inclusive?

(1) A " clean" no is an integer divisible only by 2 factors, one of which is greater than 2

(2) A " clean" no must be odd

wait a second....
6 is divisible exactly by two factors ( I assume factors don't include 1 or the number itself) 2 and 3, but it's not odd.

So we need B.

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SVP
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23 Feb 2005, 10:04
swath20 wrote:
what is the probab of selecting a " clean" number from a set of integers containing all multiples of 3 between 1 and 99, inclusive?

(1) A " clean" no is an integer divisible only by 2 factors, one of which is greater than 2

(2) A " clean" no must be odd

We are asked to find the probability. We are not asked to find one "clean" number. In other words, as some of the others already mentioned here, as long as the definition is clearly defined, then we will be able to identify all "clean" numbers and thus the probability of pick them from a clearly defined set.

On this note, (1) is clearly define, but (2) is not. (2) would be sufficient if it is formulated like this:
(2) A "clean" number means an odd number.

When it says a clean number must be odd, we do not know if there's other requirements for a clean number. Perhaps it must be odd and greater than 50? Or whatever. As long as we don't know the exact definition of a clean number, we cannot pin down the set of clean numbers, and thus cannot find the probability that we were asked to find.

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Manager
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25 Feb 2005, 11:56
Thanks Hong Hu for your detailed explanation

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Manager
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25 Feb 2005, 12:49
Should be D

As there are 33 possible outcomes or numbers.

(1) No of favorable outcomes: all Prime greater than 2 so Probab can be found

(2) favorable outcomes Only odd numbers again Probab can be found

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25 Feb 2005, 12:49
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