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If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Difficulty:   35% (medium)

Question Stats: 65% (01:40) correct 35% (01:43) wrong based on 107 sessions

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If one number is chosen at random from the first 1000 positive integers, then what is the probability that the number is a multiple of 2 and 8?

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

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If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Bunuel wrote:
If one number is chosen at random from the first 1000 positive integers, then what is the probability that the number is a multiple of 2 and 8?

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

total multiple of 2 in 1-1000 : 1000/2= 500 and of 8 , 1000/8= 125

Multiples which are of both8 & 2 are to be found since multiplayer of 8 would be of 2 as well so:
125/1000= 1/8 IMO B
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Originally posted by Archit3110 on 04 Dec 2018, 04:35.
Last edited by Archit3110 on 04 Dec 2018, 09:45, edited 1 time in total.
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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Archit3110 wrote:
Bunuel wrote:
If one number is chosen at random from the first 1000 positive integers, then what is the probability that the number is a multiple of 2 and 8?

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

total multiple of 2 in 1-1000 : 1000/2= 500 and of 8 , 1000/8= 125

500/1000 * 125/1000= 1/16 IMO A

Thing is, whatever number is a multiple of 8 is also a multiple of 2, so I would go and count only the multiples of 8 between 1 and 1000, or 125/1000 or 1/8.

Let's take a smaller range and test: between 1 and 20, there are 10 multiples of 2 and 2 multiples of 8. The answer should be 2/20 (or 1/10) and not 2/20 * 10/20, or 1/20.

Am I missing something?
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If one number is chosen at random from the first 1000 positive integer  [#permalink]

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1. Multiple of 2 from 1-1000 is 1000/2=500 => P(A)= 500/1000=1/2
2. Any multiple of 8 is also a multiple of 2 so we need to find the multiples of 8 from 1 to 1000 the first one is 8 and the last one is 1000. Simple way is 1000/8=125
=> P(B)=125/1000=1/8

P(A and B) = P(A) x P(B)=1/2*1/8=1/16
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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Since the question says,the no.is a multiple of 2 and 8,this means we need to count the multiples of 8,as every multiple of 8 is also a multiple of 2.
As there are 125 no.s between 1-1000 which are multiple of 8,so answer should be 125/1000= 1/8.

Hence B

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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Question says multiple of 2 AND 8 not OR so 1/8 is correct

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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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yoannnesme wrote:
Archit3110 wrote:
Bunuel wrote:
If one number is chosen at random from the first 1000 positive integers, then what is the probability that the number is a multiple of 2 and 8?

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

total multiple of 2 in 1-1000 : 1000/2= 500 and of 8 , 1000/8= 125

500/1000 * 125/1000= 1/16 IMO A

Thing is, whatever number is a multiple of 8 is also a multiple of 2, so I would go and count only the multiples of 8 between 1 and 1000, or 125/1000 or 1/8.

Let's take a smaller range and test: between 1 and 20, there are 10 multiples of 2 and 2 multiples of 8. The answer should be 2/20 (or 1/10) and not 2/20 * 10/20, or 1/20.

Am I missing something?

A number which would be multiple of 2 won't necessarily be a multiple of 8... But otherwise yes it would be correct to say that multiple of 8 would also be a multiple of 2.. so by doing 125/1000 you are only considering multiples of 8 only what about multiple of 2 i.e 4,6,12... which is why I calculated separate for 2 & 8 ...

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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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sumit10 wrote:
Question says multiple of 2 AND 8 not OR so 1/8 is correct

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Could you please reinstate what you mean by line "Question says multiple of 2 AND 8 not OR "??
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GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Archit3110 wrote:
sumit10 wrote:
Question says multiple of 2 AND 8 not OR so 1/8 is correct

Posted from my mobile device

Could you please reinstate what you mean by line "Question says multiple of 2 AND 8 not OR "??

Draw a two-set Venn diagram: One set is all multiples of 2. The other set is all multiples of 8. And the intersection of the two sets is what we are interested in. Is so happens that the set "Multiples of 8"is completely inside the set multiple of 2 and 8. Hence we need to find only the multiples of 8 and it comes out to be one in every 8 numbers or 1/8.

Hope this makes sense.

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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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multiple of 2 and 8 . All multiple of 8 will be multiple of 2 as well so we just need to find the multiples of 8 i.e 125
so 125/1000

should be 1/8
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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Archit3110 wrote:
sumit10 wrote:
Question says multiple of 2 AND 8 not OR so 1/8 is correct

Posted from my mobile device

Could you please reinstate what you mean by line "Question says multiple of 2 AND 8 not OR "??

Draw a two-set Venn diagram: One set is all multiples of 2. The other set is all multiples of 8. And the intersection of the two sets is what we are interested in. Is so happens that the set "Multiples of 8"is completely inside the set multiple of 2 and 8. Hence we need to find only the multiples of 8 and it comes out to be one in every 8 numbers or 1/8.

Hope this makes sense.

Best,

Okay thanks I understood ..
I actually calculated both multiples of 2 & 8...
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Re: If one number is chosen at random from the first 1000 positive integer  [#permalink]

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Bunuel wrote:
If one number is chosen at random from the first 1000 positive integers, then what is the probability that the number is a multiple of 2 and 8?

A. 1/16
B. 1/8
C. 1/4
D. 1/2
E. 2/3

total number of numbers in the set: (1000 - 1) + 1= 1000

Now let's find the subset:-
All multiples of 8 are a multiple of 2 & 8.

Between 1 and 1000
there are 125 multiples of 8. Because 8*125=1000

Probability= 125/1000.
Option B (1/8)
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I never gave up what I wanted- Re: If one number is chosen at random from the first 1000 positive integer   [#permalink] 24 May 2019, 08:06
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