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The integers from 1 to 100 inclusive are each written on a

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The integers from 1 to 100 inclusive are each written on a [#permalink] New post 09 Apr 2005, 00:13
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The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%
[Reveal] Spoiler: OA

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Re: The integers from 1 to 100 inclusive are each written on a [#permalink] New post 19 Mar 2012, 02:48
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Zem wrote:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%


# of multiples of 2 in the range (100-2)/2+1=50 (check this: totally-basic-94862.html#p730075);
# of multiples of 3 in the range (99-3)/3+1=33;
# of multiples of both 2 and 3, so multiples of 6, in the range (96-6)/6+1=16 (to get the overlap of above two sets);

Hence there are total of 50+33-16=67 numbers from 1 to 100, inclusive, which are multiples of 2 or 3, which means that the probability of selecting neither multiple of 2 nor multiple of 3 is 1-67/100=33/100=33%.

Answer: D.

Hope it's clear.
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 [#permalink] New post 09 Apr 2005, 00:44
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67%...

50/100 (all multiples of 2) + 33/100 (all multiples of 3) - 1650/10000 (multiples of 3 and 2) =6650/10000=67%...
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 [#permalink] New post 10 Apr 2005, 05:49
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100 slips of paper: 50 are even, and 17 are odd multiples of 3

100-(50+17)=33
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Re: The integers from 1 to 100 inclusive are each written on a [#permalink] New post 18 Mar 2012, 19:56
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Simple to solve using P(n) = 1-X concept

P = 1 - [ P(even 0r multiples of 2)+P(multiples of 3) - P(multiples of 2 and 3 or multiples of 6)]
P = (1 - [ 50/100+33/100 - 16/100]) = 33%
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 [#permalink] New post 09 Apr 2005, 01:39
odd multiples of 3 are 17/100
3 3+6 3+6+6 3+6+6+6 etc
50/100+17/100=67/100
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 [#permalink] New post 09 Apr 2005, 02:20
Nither even nor a multiple of 3.

Even numbers between 1 and 100 are 2, 4, 6 ... 100 (50 numbers).

Numbers that are multiple of 3 are 3, 6, 9 ... 99. However, all even multiples of 3 have been counted already in even numbers. Therefore, this set includes 3, 9, 15, 21 ... 99. (17 numbers).

Thus total numbers that are to be excluded = 50 + 17 = 67.

Therefore probability that a randomly picked number is nither even nor a multiple of 3 = 33%
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 [#permalink] New post 09 Apr 2005, 02:23
why I added instead of subtracting? :wall

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"
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 [#permalink] New post 10 Apr 2005, 02:37
thearch wrote:
why I added instead of subtracting? :wall

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"


Careful theArch, that wall is rather fragile !!!

Btw what're you saying? Your answer matches mine?
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 [#permalink] New post 10 Apr 2005, 02:46
My first answer was 67/100.
However, I was deceived by christoph, because I calculated 67/100 and then I looked at answers already given, and forgot to subtract. This wouldn't happen in the actual test :)
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 [#permalink] New post 10 Apr 2005, 05:59
thearch wrote:
why I added instead of subtracting? :wall

Options >=50% should be ruled out soon because numbers range from 1 to 100, and thus subtracting 50 even you can't have more than 49 (you know at least that 3 is an odd multiple of 3, son you have at least 1 in 50 remaining numbers)

This question is asking nothing more than:
"how many odd numbers are in the first 100 positive ones that are not multiple of 3?"


This happens with many of us....We get there to the answer and forget subtracting.......GMAT has both the answers as a trap and we should be careful enough not to fall in it........
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 [#permalink] New post 11 Apr 2005, 11:24
Zem wrote:
but how you found the number: 17?


Zem, we are trying to find the numbers that are nither divisible by 2 nor 3.

If you take the numbers that are divisible by 2 away, you've taken all even numbers between 1 and 100 away. They are (100-1)/2 = 50 numbers.

We could repeat the same process with 3. However, there are numbers that are common to 2 and 3, and we don't have to take them out (they were accounted for in the even numbers list anyway).

So we see what are these numbers we're looking at (odd multiples of 3).
They are 3, 9, 15, 21 and so on.

There're a plathora of ways to figuring out how many such numbers would exist between 1 and 100. I'd list out two here.

One is the way of sequences.

the last number in this list would be 99.
99 = 3 + (n - 1) 6 => n = 17.

Else do a 100/16 = 16.xxx. 16*6 = 96. but since it starts at 3, the 16th multiple would be 99. 16 multiples, and add one (for 3) and you've 17.

Thus we have the number 17.

Hope that helps.
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 [#permalink] New post 11 Apr 2005, 19:30
I think that in this problem it's easier to find directly the prob than to use the formula 1-(opposite prob)

from 1 to 100 : 100 numbers
even numbers : 50
multiples of 3 : 33 (3*1, 3*2, ...3*33) however in those 33 numbers there are already even numbers and you have to be careful to not repeat the even numbers you've already taken out so finally you will get 17 odd and 16 even (you can see that everytime 3 is multiply by an even number the final result is obviously even so 1/2 of the multiples of 3 will be even and 1/2 odd, howver the last one is odd so there is an additionnal odd one)

17 odd multiples of 3 + 50 even numbers = 67

100-67 = 33 :wink:
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Re: The integers from 1 to 100 inclusive are each written on a [#permalink] New post 30 Sep 2013, 10:26
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Re: The integers from 1 to 100 inclusive are each written on a [#permalink] New post 10 Dec 2013, 05:19
Zem wrote:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%


Bunuel is this approach correct?

Ballparking from 1 to 10 there are only 3 numbers that satisfy these conditions: 1,5 and 7. Then 3/10 so approximately 30% only answer choice that is close is D

If you do it for other ranges you are basically going to have 4,3,4,3,4,3 so in theory it would be something like 3.5/10 or between 30% and 35%

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Re: The integers from 1 to 100 inclusive are each written on a [#permalink] New post 10 Dec 2013, 05:22
Zem wrote:
The integers from 1 to 100 inclusive are each written on a single slip of paper and dropped into a jar. If one slip of paper is removed at random, approximately what is the probability that the number on it is neither even nor a multiple of 3?

A. 83%
B. 67%
C. 50%
D. 33%
E. 17%


Multiple of 3: 100/3 = 33 numbers
Multiple of 2: 100/2 = 50 numbers

Multiple of 6: 100/6= 16 numbers

Total numbers divisible by 2 or 3: 50 + 33 - 16 = 67

Neither = 100 - 67 = 33

Probability = 33/100

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Re: The integers from 1 to 100 inclusive are each written on a   [#permalink] 10 Dec 2013, 05:22
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