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5 integers, not necessarily distinct, are chosen from the integers

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5 integers, not necessarily distinct, are chosen from the integers [#permalink]

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New post 03 Jan 2018, 04:43
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5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive, where n is a positive integer. If the probability that the product of the chosen integers is zero is 1 – (0.9375)^5, then what is the value of n?

A) 7
B) 8
C) 9
D) 15
E) 16
[Reveal] Spoiler: OA

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Re: 5 integers, not necessarily distinct, are chosen from the integers [#permalink]

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New post 03 Jan 2018, 05:06
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saswata4s wrote:
5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive, where n is a positive integer. If the probability that the product of the chosen integers is zero is 1 – (0.9375)^5, then what is the value of n?

A) 7
B) 8
C) 9
D) 15
E) 16


15/16 is 0.9375. This means that the total number from -(n+1) to n inclusive equals 16. If 8 is the answer this means there are: (-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8) 18 numbers. Shouldn't the answer be 7?

Kudos [?]: 3 [1], given: 6

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Joined: 25 Feb 2013
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Re: 5 integers, not necessarily distinct, are chosen from the integers [#permalink]

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New post 03 Jan 2018, 05:18
mohdtaha wrote:
saswata4s wrote:
5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive, where n is a positive integer. If the probability that the product of the chosen integers is zero is 1 – (0.9375)^5, then what is the value of n?

A) 7
B) 8
C) 9
D) 15
E) 16


15/16 is 0.9375. This means that the total number from -(n+1) to n inclusive equals 16. If 8 is the answer this means there are: (-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8) 18 numbers. Shouldn't the answer be 7?


Yes agree with mohdtaha

if -(n+1) & n are inclusive, then total terms in the series=n+(n+1)+1=2n+2

and as 2n+2=16=>n=7

Kudos [?]: 383 [0], given: 42

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Re: 5 integers, not necessarily distinct, are chosen from the integers [#permalink]

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New post 03 Jan 2018, 05:51
niks18 wrote:
mohdtaha wrote:
saswata4s wrote:
5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive, where n is a positive integer. If the probability that the product of the chosen integers is zero is 1 – (0.9375)^5, then what is the value of n?

A) 7
B) 8
C) 9
D) 15
E) 16


15/16 is 0.9375. This means that the total number from -(n+1) to n inclusive equals 16. If 8 is the answer this means there are: (-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8) 18 numbers. Shouldn't the answer be 7?


Yes agree with mohdtaha

if -(n+1) & n are inclusive, then total terms in the series=n+(n+1)+1=2n+2

and as 2n+2=16=>n=7



why did u take 16


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Joined: 25 Feb 2013
Posts: 826

Kudos [?]: 383 [0], given: 42

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Re: 5 integers, not necessarily distinct, are chosen from the integers [#permalink]

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New post 03 Jan 2018, 05:58
riniglory wrote:
why did u take 16


Hi riniglory,

Refer to the earlier solution posted by mohdtaha

Total number of integers is 16 which means -(n+1) to n has 16 terms.

Kudos [?]: 383 [0], given: 42

Re: 5 integers, not necessarily distinct, are chosen from the integers   [#permalink] 03 Jan 2018, 05:58
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5 integers, not necessarily distinct, are chosen from the integers

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