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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 04 May 2019, 09:01
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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 04 May 2019, 09:48
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You might first imagine how you'd solve the problem using a concrete number. Suppose we had 9 different integers, say, and one of them was zero. Then if we pick five of them with replacement, and we want a nonzero product, we need to avoid picking zero each time. We'd have an 8/9 chance to pick a nonzero number each time, and we'd have a (8/9)^5 probability of getting a nonzero product in five selections, so we'd have a 1 - (8/9)^5 probability of getting a product of zero.

Now if we compare what we just did with the expression in the question, 1 - (0.9375)^5, we can see that 0.9375 should equal the probability that we pick a nonzero number. So if we have k numbers, (k-1)/k = 0.9375, or 1 - (1/k) = 0.9375, and rearranging, 1/k = 0.0625. If you know that 0.625 = 5/8, then 0.0625 = 5/80 = 1/16, so k = 16, and we must be picking from 16 different numbers here. Between -n-1 and n inclusive, we have n - (-n-1) + 1 = 2n + 2 numbers, so 2n + 2 = 16 and n = 7.
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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New post 14 Oct 2019, 00:52
IanStewart wrote:
You might first imagine how you'd solve the problem using a concrete number. Suppose we had 9 different integers, say, and one of them was zero. Then if we pick five of them with replacement, and we want a nonzero product, we need to avoid picking zero each time. We'd have an 8/9 chance to pick a nonzero number each time, and we'd have a (8/9)^5 probability of getting a nonzero product in five selections, so we'd have a 1 - (8/9)^5 probability of getting a product of zero.

Now if we compare what we just did with the expression in the question, 1 - (0.9375)^5, we can see that 0.9375 should equal the probability that we pick a nonzero number. So if we have k numbers, (k-1)/k = 0.9375, or 1 - (1/k) = 0.9375, and rearranging, 1/k = 0.0625. If you know that 0.625 = 5/8, then 0.0625 = 5/80 = 1/16, so k = 16, and we must be picking from 16 different numbers here. Between -n-1 and n inclusive, we have n - (-n-1) + 1 = 2n + 2 numbers, so 2n + 2 = 16 and n = 7.


IanStewart can u pls tell me why did u tell that we would pick five numbers out of nine with REPLACEMENT? Why should we assume that the taken numbers will be replaced?
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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New post 14 Oct 2019, 11:09
Hea234ven wrote:
IanStewart can u pls tell me why did u tell that we would pick five numbers out of nine with REPLACEMENT? Why should we assume that the taken numbers will be replaced?


Because the question says: "5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive". But you wouldn't see wording like this on the real test -- on the real test they'd say "with replacement".
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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New post 14 Oct 2019, 21:50
nidhi

start with a simple example, What is the probability for the number 0,1,2 such that when chosen 5 times, with replacement, the product is non zero?

it is \frac{2}{3}*2/3*2/3*2/3*2/3.

If zero then 1- the before equation.

so we can say if .9375^5 is zero then .0625^5 is non zero. or (625/10000)^5
after deducting (1/16)^5 now we know that we have to choose between 16 numbers and since n is positive and we have to go upto -(n+1) start checking with numbers
if n is 7 we get 16 number from 7 to -8. for others the number will be higher. Thus A.
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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New post 15 Oct 2019, 02:34
IanStewart wrote:
Hea234ven wrote:
IanStewart can u pls tell me why did u tell that we would pick five numbers out of nine with REPLACEMENT? Why should we assume that the taken numbers will be replaced?


Because the question says: "5 integers, not necessarily distinct, are chosen from the integers between –(n+1) and n, inclusive". But you wouldn't see wording like this on the real test -- on the real test they'd say "with replacement".


That's exactly where I got stuck.

Whether to take probability or non zero as
\(\frac{(2n+1)^5}{(2n+2)^5}\) OR \(\frac{(2n+1)*(2n)*(2n-1)*(2n-2)*(2n-3)}{(2n+2)^5}\)

Anyway it does not matter as I marked B which is wrong.
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet  [#permalink]

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New post 15 Oct 2019, 07:44
There a couple of clues that are really important here.

One, the fact that we have integers stretching from the negative end of the spectrum to the positive. Two, the fact that we have to find the probability that the product of the chosen integers is ZERO.
The product of any set of integers can be ZERO only when there’s at least one ZERO in the set. The fact that we have both negative and positive integers is enough to tell us that ZERO can be in the set of chosen numbers.

When I looked at 0.9375, the first thing I did was to subtract it from 1; it gave me 0.0625, but more importantly it helped me figure out that 0.0625 is \(\frac{1}{16}\).

Finding the probability that the product of the chosen integers is ZERO is rather difficult, this is because we do not know how many of the integers are ZERO. On the contrary, finding out the probability that the product is NOT ZERO is easy; because we know that NONE of the integers can be ZERO.

Therefore,
Probability (Product of chosen integers being ZERO) = 1 – (Product of chosen integers not ZERO).

Fortunately, that’s what is given to us as 1 – \((0.9375)^5\). If 0.0625 = \(\frac{1}{16}\), then 0.9375 = \(\frac{15}{16}\). So, we have the probability of chosen integers not being zero as \((\frac{15}{16})^5\). This can only happen when there are 16 integers in total and one of them is ZERO.

This means, there need to be a total of 16 integers between –(n+1) and n. Plugging in the value of 7, we can calculate that there are a total of 16 integers between -8 and 7, inclusive. Hence, 7 has to be the answer.
The correct answer option is A.

In probability questions, P ( E ) = 1 – P ( E’) is a very useful concept if you know the right way of incorporating it into your solution.

Hope that helps!
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Re: 5 integers, not necessarily distinct, are chosen from the integers bet   [#permalink] 15 Oct 2019, 07:44
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