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What is the probability of getting a sum of 12 when rolling

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What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post Updated on: 23 Feb 2014, 23:58
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What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?

A. 10/216
B. 12/216
C. 21/216
D. 23/216
E. 25/216

Originally posted by anhlukas on 16 Aug 2006, 12:22.
Last edited by Bunuel on 23 Feb 2014, 23:58, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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New post 16 Aug 2006, 15:11
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Sum of 12 can be achieved in following ways
6,5,1---Total cases = 3! = 6
6,4,2---Total cases = 3!= 6
6,3,3---Total cases = 3!/2! = 3
5,5,2---Total cases = 3!/2! = 3
5,4,3---Total cases = 3! = 6
4,4,4---Total cases = 3!/3! = 1

Total cases = 25

Probability = 25 * (1/6 * 1/6 * 1/6) = 25/216
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New post 16 Aug 2006, 13:41
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I broke my head 15 minutes on it, then the soln is so devilishly simple! :wall

Answer is E > 25/100

Basically, if the numbers on two of the dice are decided, then there is only one value of the third die that is possible. So you have to find only those combinations of two dice that yield a sum of 6 or above. e.g. If the value on the 1st die is 1, then you have two possible outcomes - 5 or 6 on the 2nd die. Similarly, for a value of 2 on the first die, you have 3 possible outcomes (4, 5, 6) and so on... However, if you get a value of 6 on the first die, then you can have only 5 possible outcomes.

The total number of outcomes comes out to be (2 + 3 + 4 + 5 + 6 + 5) = 25.

Probability is 25/216!
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Re: What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 23 Mar 2015, 04:47
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anhlukas wrote:
What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?

A. 10/216
B. 12/216
C. 21/216
D. 23/216
E. 25/216



let the dices be a,b,c

a+b+c = 12

but 0 < a,b,c < 6

so we can re-write the equation as

(6-a)+(6-b)+(6-c) = 12
=> a+b+c = 6
which has total of C(8,2) whole number solutions = 28
but out of these 28, three cases must be where 2 of a,b, or c is 0 and the other is 6. so we need to remove those cases which would be 3 cases ( C(3,2) )
so total 25 such cases

answer = 25/216.
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Re: What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 04 Apr 2015, 13:48
I think it is very easy...........
Start checking from the smaller or bigger numbers on the dice. We will check from bigger numbers working downwards: start with 6, it has the following options: (6,5,1), (6,4,2), (6,3,3). Now pass on to 5: (5,5,2), (5,4,3). Now 4: (4,4,4). And that’s it, these are all number combinations that are possible, if you go on to 3, you will notice that you need to use 4, 5 or 6, that you have already considered (the same goes for 2 and 1). Now analyze every option: 6,5,1 has 6 options (6,5,1), (6,1,5), (5,1,6), (5,6,1), (1,6,5), (1,5,6). So do (6,4,2) and (5,4,3). Options (6,3,3) and (5,5,2) have 3 options each: (5,5,2), (5,2,5) and (2,5,5). The same goes for (6,3,3). The last option (4,4,4) has only one option. The total is 3*6+2*3+1=18+6+1 = 25 out of 216 (6x6x6) options.

5,5,2= (552),(255),(525)
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What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 25 Aug 2015, 13:56
An easy way to visualise it is by drawing a Binomial tree, we've all done those in school and if each branch has a 1/2 probability, you'll find that there's only 3 ways to get there.

(1/2)^3 x 3 = 3/8


Hope it helps!

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Re: What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 25 Aug 2015, 22:25
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anhlukas wrote:
What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?

A. 10/216
B. 12/216
C. 21/216
D. 23/216
E. 25/216



Check out these two posts:

http://www.veritasprep.com/blog/2012/10 ... l-picture/
http://www.veritasprep.com/blog/2012/10 ... e-part-ii/

They discuss a shortcut to "how to find the probability of obtaining a given sum when 3 dice are rolled"
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Re: What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 20 Sep 2015, 11:05
VeritasPrepKarishma wrote:
Check out these two posts:

http://www.veritasprep.com/blog/2012/10 ... l-picture/
http://www.veritasprep.com/blog/2012/10 ... e-part-ii/

They discuss a shortcut to "how to find the probability of obtaining a given sum when 3 dice are rolled"



Excellent!!! Thanks Karishma
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Re: What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 24 Nov 2015, 06:59
Thanks Karishma,

Could you please explain how 9 is calculated that is subtracted from 9C2 in below scenario?

Similarly, how will you adjust for the sum of 10? There will be cases where the split is like this: (7, 0, 0) or (6, 1, 0). These do not work since the maximum a die can show is 6. So you need to remove 9 cases (3 arrangements of 7, 0, 0 and 6 arrangements of 6, 1, 0)
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Re: What is the probability of getting a sum of 12 when rolling  [#permalink]

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New post 09 Jul 2017, 02:53
thefibonacci wrote:
anhlukas wrote:
What is the probability of getting a sum of 12 when rolling 3 dice simultaneously?

A. 10/216
B. 12/216
C. 21/216
D. 23/216
E. 25/216



let the dices be a,b,c

a+b+c = 12

but 0 < a,b,c < 6

so we can re-write the equation as

(6-a)+(6-b)+(6-c) = 12
=> a+b+c = 6
which has total of C(8,2) whole number solutions = 28
but out of these 28, three cases must be where 2 of a,b, or c is 0 and the other is 6. so we need to remove those cases which would be 3 cases ( C(3,2) )
so total 25 such cases

answer = 25/216.

I understand using this method and have used. But here I didn't understand the rewriting the equation part. Please explain that and what follows that. Bunuel, if possible, would be great if you could help me out here.
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Re: What is the probability of getting a sum of 12 when rolling   [#permalink] 09 Jul 2017, 02:53
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