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# What is the probability of getting a sum of 8 or 14 when

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Director
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What is the probability of getting a sum of 8 or 14 when [#permalink]  26 Nov 2004, 06:31
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Question Stats:

30% (03:53) correct 70% (02:54) wrong based on 182 sessions
What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

A. 1/6
B. 1/4
C. 1/2
D. 21/216
E. 32/216
[Reveal] Spoiler: OA
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]  26 Nov 2004, 06:51
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Quote:
"Gayathri"
What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

a) 1/6
b) 1/4
c) 1/2
d) 21/216
e) 32/216

Is there a shorter way to do this than listing each possibility?

I am confident that there is a better way, but I looked at the first few possibilities

throw a 3 - 1 way
throw a 4 - 3 ways
throw a 5 - 6 ways
throw a 6 - 10 ways

and recognized these as triangular numbers (the series could also be identified as add 2, add 3, add 4 ...).

from there it was easy to calculate that there are 21 ways to hit an 8 and 15 ways to hit a 14. (Problems with fair dice and coins produce symmetrical probability distributions, so one can count down from 1 way to throw an 18)

[more than 2 minutes - less than 3 minutes]
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]  26 Nov 2004, 22:03
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Richie Weasel wrote:
Quote:
"Gayathri"
What is the probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously?

a) 1/6
b) 1/4
c) 1/2
d) 21/216
e) 32/216

Is there a shorter way to do this than listing each possibility?

I am confident that there is a better way, but I looked at the first few possibilities

throw a 3 - 1 way
throw a 4 - 3 ways
throw a 5 - 6 ways
throw a 6 - 10 ways

and recognized these as triangular numbers (the series could also be identified as add 2, add 3, add 4 ...).

from there it was easy to calculate that there are 21 ways to hit an 8 and 15 ways to hit a 14. (Problems with fair dice and coins produce symmetrical probability distributions, so one can count down from 1 way to throw an 18)

[more than 2 minutes - less than 3 minutes]

Wow, Richie, this is a fast and easy way to solve it! I will take a note of this method!

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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]  27 Nov 2004, 22:06
Is there any other to solve this question ?

Can you elaborate the answer pls. ?
Director
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]  28 Nov 2004, 05:05
gmat+obsessed wrote:
Is there any other to solve this question ?

The other way would be to list out each option.
The options for a sum of 8: (6,1,1) has 3 options ie (6,1,1), (1,6,1), (1,1,6); (5,2,1) has 6 options, (4,3,1) has 6 options, (4,2,2) has 3 options, (3,3,2) has 3 options. We have 21 options to get 8.

The options for a sum of 14: (6,4,4) has 3 options, (6,5,3) has 6 options, (6,6,2) has 3 options, (5,5,4) has 3 options. We have 15 options to get 14.

Total: 21+15= 36/216 = 1/6.
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]  12 Sep 2014, 05:45
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]  22 Oct 2014, 04:17
I missed some options for 8 as a sum. Idea is that options containing three different number have 3!=6 possible events, options containing 2 identical and 1 other have 3!/2!*1!=3 possible events. By multiplying and summing we get the answer
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What is the probability of getting a sum of 8 or 14 when [#permalink]  17 Nov 2014, 13:03
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Here's another way:
You start by listing out each possible triplet for every number on the dice for each of the two numbers we need (notice that for each triplet the probability is $$\frac{1}{6^3}$$):
8:
1:
[1,1,6]
[1,2,5]
[1,3,4]
[1,4,3]
[1,5,2]
[1,6,1]

2:
[2,1,5]
[2,2,4]
[2,3,3]
[2,4,2]
[2,5,1]

…and so on...

Soon thereafter, you realize that, for each number of the dice, there is one triplet less that adds up to 8, than the previous number (i.e. 1-triplets: 6, 2-triplets: 5, 3-triplets: 4, and so on…). Given that $$\frac{1}{6^3}$$ is a common factor to all triplets, we get that: $$\frac{1}{6^3}*(6+5+4+3+2+1)$$ or $$\frac{1}{6^3}*(21)$$.

Then we do a similar process for 14:

1:
(No possible combination adds up to 14)

2:
[2,6,6]

3:
[3,5,6]
[3,6,5]

…and so on…

So you'll notice that a similar thing happens in this case: from 2 on, for each number of the dice, there is one triplet more that adds up to 14, than the previous number (i.e. 2-triplet: 1, 3-triplets: 2, 4-triplets: 3, and so on…). Again, given that $$\frac{1}{6^3}$$ is a common factor to all triplets, we get that: $$\frac{1}{6^3}*(1+2+3+4+5)$$ or $$\frac{1}{6^3}*(15)$$.

Given that we need the probability of getting a sum of 8 OR 14, we add up both of these cases:
$$\frac{1}{6^3}*(21)+\frac{1}{6^3}*(15)$$

We factor out $$\frac{1}{6^3}$$, and find that:
$$\frac{1}{6^3}*(36)$$, and by recognizing that $$36 = 6^2$$, we cross it out with $$6^3$$ to find that the probability of getting a sum of 8 or 14 when rolling three fair dice is $$\frac{1}{6}$$.

What is the probability of getting a sum of 8 or 14 when   [#permalink] 17 Nov 2014, 13:03
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