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

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PriyankaPalit7
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5 integers, not necessarily distinct, are chosen from the integers bet [#permalink] New post   21 Mar 2019, 09:40
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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

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chetan2u
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet [#permalink] New post   21 Mar 2019, 20:48
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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
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet [#permalink] New post   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


Thanks for your explanation Chetan.
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?
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet [#permalink] New post   22 Mar 2019, 09:20
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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


Thanks for your explanation Chetan.
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.

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet [#permalink]
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