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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]

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16 Oct 2006, 22:27
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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%
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16 Oct 2006, 23:28
My pick is E...

Number of even intergers: 50

Number of intergers that are the multiples of 3: around 100/3=33

So the probability that an even interger or an interger multiple of 3 has been removed is:

50/100+33/100=1/2+1/3=5/6

The probability that neither of those numbers have been removed:

1-5/6=1/6=17%
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16 Oct 2006, 23:54
It must be 33%

Total multiples of 2 = 50
Total multiples of 3 = 33
Now the numbers which are divisible by both 2 and 3 i.e numbers which are divisible by 6 are counted in the first 50 numbers as well as in the next 33 numbers. So we have to exclude the numbers that are divisible by 6 once from 50+33 =83

So total multiples of 6 =16

So 83-16 =67.

So there are 67 numbers which are either even or multiples of 3.

So the prob is 33/100 = 33%
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17 Oct 2006, 07:16
OA is D
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17 Oct 2006, 13:01
Yes D. There are 50 even numbers from 1-100.

There are 17 numbers that are multiplies of three that are not even:
3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99.
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Why this is not true [#permalink]

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17 Oct 2006, 14:46
If we negate this, we get "Even and multiple of 3", we have 16 numbers
(6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96). So this comes to 16/100. So for the answer (1-16/100=84/100). So why not 83% approx is the answer.I am just confused.Can any one correct me?
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17 Oct 2006, 15:51
It isn't even AND a multiple of 3.....its even NOR a multiple of 3
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