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Manager  Joined: 08 Sep 2010
Posts: 150
Location: India
WE 1: 6 Year, Telecom(GSM)
If four dice are thrown together , then the probability that the sum  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 29% (01:08) correct 71% (02:13) wrong based on 19 sessions

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If four dice are thrown together , then the probability that the sum on them together is either 19 or 23

1. 4/108
2. 12/108
3. 9/108
4. 5/108
5. 6/108

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Intern  Joined: 26 Mar 2010
Posts: 26
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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Is the answer 5/108? Just want to make sure so that I don't explain a wrong answer and confuse people...
Intern  Joined: 30 Sep 2010
Posts: 47
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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3
no of ways to get 19 = no of ways to get 9 = C(8,5) = 56
no of ways to get 23 = no of ways to get 5 = C(4,1) = 4

So total ways = 60
probability = 60/1296 = 5/108 (I need kudos)

Find the simpler way

Dice problem

A nice way to find the sum:

e.g 4 dices are thrown

then the outcome of the sums can be:

4,5,6,………………14……………..,22,23,24

Mid point is 14. The left side of the mid point is mirror image of the right part.

So lets say the probability of getting a 20 is same as the probability of getting a 8

In this case the number of ways to get the sums are:

4: C(3,0) = 1 (start 1 before the # of throws)
5: C(4,1) = 4
6: C(5,2) = 10
7: C(6,3) = 20
8: C(7,4) = 35

So probability of getting a 20 = probability of getting a 8 = 35/6^4 = 35/1296
Similarly for rolloing of 3 dices:

then the outcome of the sums can be:
Mid point is 10.5. The left side of the mid point is mirror image of the right part.

3,4,5,6,………………10.5……………..,15.16.17.18

In this case the number of ways to get the sums are:

3: C(2,0) = 1 (start 1 before the # of throws)
4: C(3,1) = 3
5: C(4,2) = 6
6: C(5,3) = 10
7: C(6,4) = 15
8: C(7,5) = 21
Manager  Joined: 27 Mar 2010
Posts: 77
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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krushna wrote:
no of ways to get 19 = no of ways to get 9 = C(8,5) = 56
no of ways to get 23 = no of ways to get 5 = C(4,1) = 4

So total ways = 60
probability = 60/1296 = 5/108 (I need kudos)

Find the simpler way

Dice problem

A nice way to find the sum:

e.g 4 dices are thrown

then the outcome of the sums can be:

4,5,6,………………14……………..,22,23,24

Mid point is 14. The left side of the mid point is mirror image of the right part.

So lets say the probability of getting a 20 is same as the probability of getting a 8

In this case the number of ways to get the sums are:

4: C(3,0) = 1 (start 1 before the # of throws)
5: C(4,1) = 4
6: C(5,2) = 10
7: C(6,3) = 20
8: C(7,4) = 35

So probability of getting a 20 = probability of getting a 8 = 35/6^4 = 35/1296
Similarly for rolloing of 3 dices:

then the outcome of the sums can be:
Mid point is 10.5. The left side of the mid point is mirror image of the right part.

3,4,5,6,………………10.5……………..,15.16.17.18

In this case the number of ways to get the sums are:

3: C(2,0) = 1 (start 1 before the # of throws)
4: C(3,1) = 3
5: C(4,2) = 6
6: C(5,3) = 10
7: C(6,4) = 15
8: C(7,5) = 21

Hi, thanks for the solution

But how u got no of ways to get 9 = C(8,5), what i don't understand is how to think about c(8,5) quickly???
for sum of 4(in case of 4 dices) i can understand that no of ways=1 because each dice=1 then sum=4. but didn't understand why have to written it as c(3,0)???
Intern  Joined: 30 Sep 2010
Posts: 47
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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the formula I came up with is based on the pattern of the number of ways.
I mean I first got the answers in the normal way(by calculating all the possibilities) and then I noticed a certain pattern and finally devised this formula. You can think, its my PATENT but I am sure this works for everything. I explained in my example about the pattern. This will help you to get the answer to any SUM problem in 1 minute. GMAT is all about saving time.

If you have problem in understanding the pattern, shoot another question and I will answer it with the formula
Manager  Joined: 27 Mar 2010
Posts: 77
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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krushna wrote:
the formula I came up with is based on the pattern of the number of ways.
I mean I first got the answers in the normal way(by calculating all the possibilities) and then I noticed a certain pattern and finally devised this formula. You can think, its my PATENT but I am sure this works for everything. I explained in my example about the pattern. This will help you to get the answer to any SUM problem in 1 minute. GMAT is all about saving time.

If you have problem in understanding the pattern, shoot another question and I will answer it with the formula

Thanks a lot, I got it... +1 to you Retired Moderator Joined: 02 Sep 2010
Posts: 726
Location: London
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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If you are curious where these results come from, read : http://mathworld.wolfram.com/Dice.html

This is well beyond the scope of the exam though. And personally I am not a proponent of remembering too many formulae.

Posted from my mobile device
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Intern  Joined: 08 May 2010
Posts: 10
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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What is the source of this question?

Could this really be a GMAT question?
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9706
Location: Pune, India
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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sjgudapa wrote:
What is the source of this question?

Could this really be a GMAT question?

You could get a question such as 'If four dice are thrown together , then the probability that the sum on them is 23 is .....'
This is a simple question where you do not need to do laborious calculations. You just need to use logic. The chances of the original question featuring in GMAT are pretty slim. Remember, you need to solve a question in under two minutes on average. And they definitely do not expect you to remember formulas other than the standard ones such as area of circle etc.
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Joined: 24 Jun 2008
Posts: 1806
Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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krushna wrote:
You can think, its my PATENT You can't patent something that de Moivre came up with 270 years ago To solve general 'dice sum' problems, mathematicians use what are known as 'generating functions'. Those are miles and miles beyond the scope of the GMAT. Dice sums are generally very awkward to work out except in special cases. As Karishma pointed out, it's not hard to count the number of ways to get a sum of 23, because that's an extreme case - you must get three 6's and a 5. To count the number of ways to get a 19 is much more time-consuming, however, and is definitely not the type of thing you'll ever need to do on the GMAT.

It's very unlikely you'd even be asked the probability of getting a sum of 23 on the GMAT, not because the question is too difficult, but rather because the question is too 'boring'; you can only do it by 'brute force' enumeration. Most GMAT counting/probability questions contain some kind of twist, trick or shortcut that makes them more 'interesting' than your standard 'plug numbers into a formula' question that any spreadsheet could solve. That's one of the reasons I completely agree with shrouded's caution above about memorizing formulas - the GMAT is simply not a test of how many formulas you can remember.
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Re: If four dice are thrown together , then the probability that the sum  [#permalink]

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ankitranjan wrote:
If four dice are thrown together , then the probability that the sum on them together is either 19 or 23

1. 4/108
2. 12/108
3. 9/108
4. 5/108
5. 6/108

Notice that the probability of getting a sequence of 4 numbers in a specific order, e.g., 1-2-3-4, is 1/6 x 1/6 x 1/6 x 1/6 = 1/6^4.

Now let’s list all the groups of 4 numbers that add up to 19 with the number of ways the numbers within a group can be arranged in parenthesis:

6-6-6-1 (4!/3! = 4)

6-6-5-2 (4!/2! = 12)

6-6-4-3 (4!/2! = 12)

6-5-5-3 (4!/2! = 12)

6-5-4-4 (4!/2! = 12)

5-5-5-4 (4!/3! = 4)

Thus the probability of getting a sum of 19 is (4 x 2 + 12 x 4) x 1/6^4 = 56/6^4.

Similarly, let’s list all the groups of 4 numbers that add up to 23 with the number of ways the numbers within a group can be arranged in parenthesis:

6-6-6-5 (4!/3! = 4)

We see that there is only one group of 4 numbers that add up to 23 and within this group, there are 4 ways to arrange them. Thus the probability of getting a sum of 23 is 4 x 1/6^4 = 4/6^4.

Thus the probability of getting a sum of 19 or 23 is 56/6^4 + 4/6^4 = 60/6^4 = (6 x 10)/6^4 = 10/6^3 = 10/216 = 5/108.

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