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

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

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

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

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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]

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New post 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


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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Re: 5 integers, not necessarily distinct, are chosen from the integers bet   [#permalink] 22 Mar 2019, 09:20
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