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A bag contains six tokens, each labeled with one of the inte

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A bag contains six tokens, each labeled with one of the inte [#permalink]

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A
B
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D
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A bag contains six tokens, each labeled with one of the integers from 1 to 6, inclusive. Each integer appears on one token. When three tokens are removed at random, without replacement, what is the probability that the sum of numbers on those three tokens is equal to the positive integer N?

(1) There is only one combination of three tokens in the bag that sums to N
(2) N is a multiple of 7

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No explanation provided - Need help in solving.
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[Reveal] Spoiler: OA

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Re: A bag contains 6 tokens. [#permalink]

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New post 27 Nov 2013, 23:12
Actually, I think A is a better answer than D.

Reasons:
Statement 1) Whatever N is, there is only one combination to give it. N cannot be all 6,7,14 and 15. So probability has to be 1/20. Sufficient.

Statement 2) What if N = 7000. In that case answer is 0. Insufficient.

Edit: Tough one.. I thought the answer was an easyD
A...
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Re: A bag contains 6 tokens. [#permalink]

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New post 27 Nov 2013, 23:24
Probability of getting sum of N from 3 tokens= no. of combinations from 3 tokens which leads to N/ total sum of combination from 3 token
1)---> only 1 combnination of exists which leads to N... Hence, total outcomes =20.. using 6C3... 1/20..
2) N is a multiple of 7.. Two such combinations exist.. 1,2,4 = 7 and 3,6,5 = 14... Probabilty = 2/20


Both the statements seem to contradict each other

What is OA ???

Dont know what am I missing here

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Re: A bag contains 6 tokens. [#permalink]

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perseverance84 wrote:
Probability of getting sum of N from 3 tokens= no. of combinations from 3 tokens which leads to N/ total sum of combination from 3 token
1)---> only 1 combnination of exists which leads to N... Hence, total outcomes =20.. using 6C3... 1/20..
2) N is a multiple of 7.. Two such combinations exist.. 1,2,4 = 7 and 3,6,5 = 14... Probabilty = 2/20


Both the statements seem to contradict each other

What is OA ???

Dont know what am I missing here


The OA is B. I just did an excel workpaper to solve this problem and it appears there are four combinations that result in one number. It is 1,2,3 (sum=6), 1, 2, 4 (sum=7), 3,5,6 (sum= 14) and 4,5,6 (sum= 15). Here is the whole chart of the 20 combinations.

1 2 3 6 One combination
1 2 4 7 One combination
1 2 5 8
1 2 6 9
1 3 4 8
1 3 5 9
1 3 6 10
1 4 5 10
1 4 6 11
1 5 6 12
2 3 4 9
2 3 5 10
2 3 6 11
2 4 5 11
2 4 6 12
2 5 6 13
3 4 5 12
3 4 6 13
3 5 6 14 One combination
4 5 6 15 One combination

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Re: A bag contains 6 tokens. [#permalink]

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New post 27 Nov 2013, 23:50
Edited my previous answer. I go with A. Still don't agree with the OA though..
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Re: A bag contains six tokens, each labeled with one of the inte [#permalink]

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New post 28 Aug 2017, 23:44
Bunuel, please help in this question. I am unable to get the question stem.
Does it not mean we just have 6 tokens with 1-6 written on it without duplicates?
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Re: A bag contains six tokens, each labeled with one of the inte [#permalink]

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RMD007 wrote:
Bunuel, please help in this question. I am unable to get the question stem.
Does it not mean we just have 6 tokens with 1-6 written on it without duplicates?


Yes, it means that there are 6 tokens numbered from 1 to 6, inclusive: [1]; [2]; [3]; [4]; [5]; [6].

A bag contains six tokens, each labeled with one of the integers from 1 to 6, inclusive. Each integer appears on one token. When three tokens are removed at random, without replacement, what is the probability that the sum of numbers on those three tokens is equal to the positive integer N?

Notice several things:
1. We are picking without replacement, so if we pick [1] there won't be any more [1]'s left in the bag.
2. The least possible sum (N) is 1 + 2 + 3 = 6 and the greatest possible sum is 6 + 5 + 4 = 15.
3. The sums from 6 to 15 has different ways (combinations) to occur. For example, we can get 6 only in one way: 1 + 2 + 3 = 6 but we can get 8 in 2 ways: 1 + 2 + 5 = 1 + 3 + 4 = 8.
4.We can get the number of combinations for each sum, thus we can get the total number of combinations for all sums.

(1) There is only one combination of three tokens in the bag that sums to N. There are many value of N possible for example, if N = 1, then the probability is 0 (you cannot get the sum of 1 by adding three positive integers). But if say N = 6 (we can get 6 in one way: 6 = 1 + 2 + 3 only), then the probability is not 0 (it does not matter what it's actually is). Not sufficient.

(2) N is a multiple of 7. The greatest sum is 6 + 5 + 4 = 15, so for N to be a multiple of 7 it must be 7 or 14. Both of these numbers can be broken into the sum of three distinct positive integers only in one way: 7 = 1 + 2 + 4 and 14 = 6 + 5 + 3. Sufficient.

Answer: B.

Hope it helps.
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New post 29 Aug 2017, 00:55
the answer should not be B
because there is more than 1 combination has a multiple of 7.
waiting someone to explain on this, thank you so much

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New post 29 Aug 2017, 01:44
pclawong wrote:
the answer should not be B
because there is more than 1 combination has a multiple of 7.
waiting someone to explain on this, thank you so much


The answer IS B and it's explained in the post just above yours.

For (2): it does not matter whether N is 7 or 14, the probability will still be the same for either of these values.
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Re: A bag contains six tokens, each labeled with one of the inte [#permalink]

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New post 29 Aug 2017, 03:38
Hello Bunuel,

Can you please explain more on this,

Bunuel wrote:

(1) There is only one combination of three tokens in the bag that sums to N. There are many value of N possible for example, if N = 1, then the probability is 0 (you cannot get the sum of 1 by adding three positive integers). But if say N = 6 (we can get 6 in one way: 6 = 1 + 2 + 3 only), then the probability is not 0 (it does not matter what it's actually is). Not sufficient.


A mentions that "there is only one combination of three tokens in the bag that sums to N", so how N=1 become one possibility.

Please help !!
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Re: A bag contains six tokens, each labeled with one of the inte [#permalink]

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godot53 wrote:
Hello Bunuel,

Can you please explain more on this,

Bunuel wrote:

(1) There is only one combination of three tokens in the bag that sums to N. There are many value of N possible for example, if N = 1, then the probability is 0 (you cannot get the sum of 1 by adding three positive integers). But if say N = 6 (we can get 6 in one way: 6 = 1 + 2 + 3 only), then the probability is not 0 (it does not matter what it's actually is). Not sufficient.


A mentions that "there is only one combination of three tokens in the bag that sums to N", so how N=1 become one possibility.

Please help !!


Yes, you can get N = 1 in only one way (in 0 ways) but the probability of getting the sum of 1 is 0, while you can get N = 6 also in only one way (in 1 way) but the probability of getting the sum of 6 is NOT 0.
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Re: A bag contains six tokens, each labeled with one of the inte [#permalink]

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New post 13 Oct 2017, 00:50
I am a little lost here.

"what is the probability that the sum of numbers on those tokens is equal to the positive integer N"
"the positive integer N". i might be mistake but "the number" should be one specific number.
the probability of getting 7= the probability of getting 14. i am perfectly fine here.
but probability of getting either 7 or 14 is twice as much.
let us imagine this is not a DS question and it is a problem solving assignment where we have the general statement + the 2th (B) statement, "what is the probability that the sum of numbers on those tokens is equal to the positive integer N"

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Re: A bag contains six tokens, each labeled with one of the inte   [#permalink] 13 Oct 2017, 00:50
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