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

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Manager
Joined: 28 May 2018
Posts: 150
Location: India
Schools: ISB '21 (A)
GMAT 1: 640 Q45 V35
GMAT 2: 670 Q45 V37
GMAT 3: 730 Q50 V40
5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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21 Mar 2019, 09:40
6
00:00

Difficulty:

95% (hard)

Question Stats:

25% (02:55) correct 75% (02:53) wrong based on 20 sessions

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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
Math Expert
Joined: 02 Aug 2009
Posts: 8281
Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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21 Mar 2019, 20:48
PriyankaPalit7 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

Probability woul ddepend on total number of integers, so n+1 negative integers, n positive integers and a 0.. Total n+1+n+1=2n+2..

Now we have 1 zero and 2n+1 other integers.
It is easy to find the probability of product not being 0, so let us find that.
so we can choose the 5 integers in (2n+1)*(2n+1).. = $$(2n+1)^5$$
Probability = $$\frac{(2n+1)^5}{(2n+2)^5}$$..

Hence the probability of the product to be 0 = 1-$$\frac{(2n+1)^5}{(2n+2)^5}$$=$$1 – (0.9375)^5$$
Thus $$\frac{(2n+1)^5}{(2n+2)^5}$$=$$(0.9375)^5$$ or $$\frac{(2n+1)}{(2n+2)}$$=$$(0.9375)$$...
2n+1=(2n+2)(0.9375)=1.875n+1.875.......0.125n=0.875 or n=7

A
_________________
Manager
Joined: 28 May 2018
Posts: 150
Location: India
Schools: ISB '21 (A)
GMAT 1: 640 Q45 V35
GMAT 2: 670 Q45 V37
GMAT 3: 730 Q50 V40
Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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22 Mar 2019, 09:13
chetan2u wrote:
PriyankaPalit7 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

Probability woul ddepend on total number of integers, so n+1 negative integers, n positive integers and a 0.. Total n+1+n+1=2n+2..

Now we have 1 zero and 2n+1 other integers.
It is easy to find the probability of product not being 0, so let us find that.
so we can choose the 5 integers in (2n+1)*(2n+1).. = $$(2n+1)^5$$
Probability = $$\frac{(2n+1)^5}{(2n+2)^5}$$..

Hence the probability of the product to be 0 = 1-$$\frac{(2n+1)^5}{(2n+2)^5}$$=$$1 – (0.9375)^5$$
Thus $$\frac{(2n+1)^5}{(2n+2)^5}$$=$$(0.9375)^5$$ or $$\frac{(2n+1)}{(2n+2)}$$=$$(0.9375)$$...
2n+1=(2n+2)(0.9375)=1.875n+1.875.......0.125n=0.875 or n=7

A

However, I have one doubt.
The question statement mentions that the numbers are not necessarily distinct. Then how can we assume that there is only 1 zero?
Math Expert
Joined: 02 Aug 2009
Posts: 8281
Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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22 Mar 2019, 09:20
1
PriyankaPalit7 wrote:
chetan2u wrote:
PriyankaPalit7 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

Probability woul ddepend on total number of integers, so n+1 negative integers, n positive integers and a 0.. Total n+1+n+1=2n+2..

Now we have 1 zero and 2n+1 other integers.
It is easy to find the probability of product not being 0, so let us find that.
so we can choose the 5 integers in (2n+1)*(2n+1).. = $$(2n+1)^5$$
Probability = $$\frac{(2n+1)^5}{(2n+2)^5}$$..

Hence the probability of the product to be 0 = 1-$$\frac{(2n+1)^5}{(2n+2)^5}$$=$$1 – (0.9375)^5$$
Thus $$\frac{(2n+1)^5}{(2n+2)^5}$$=$$(0.9375)^5$$ or $$\frac{(2n+1)}{(2n+2)}$$=$$(0.9375)$$...
2n+1=(2n+2)(0.9375)=1.875n+1.875.......0.125n=0.875 or n=7

A

However, I have one doubt.
The question statement mentions that the numbers are not necessarily distinct. Then how can we assume that there is only 1 zero?

'The numbers are not necessarily distinct' is meant for picking of numbers, so you can pick up 0 5 times or 4 times with some other integer once and so on.
However numbers -(n+1) to n means all integer that are in this range. so -(n+1), -n, -(n-1),......-3, -2, -1, 0, 1, 2, 3.......n-1, n..... so each integer, including 0, is present once only.
_________________
Re: 5 integers, not necessarily distinct, are chosen from the integers bet   [#permalink] 22 Mar 2019, 09:20
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