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What is the probability of getting a sum of 8 or 14 when [#permalink]
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26 Nov 2004, 07:31
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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
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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26 Nov 2004, 07:51
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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]
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26 Nov 2004, 23:03
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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]
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27 Nov 2004, 23:06
Is there any other to solve this question ?
Can you elaborate the answer pls. ?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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28 Nov 2004, 06:05
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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]
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22 Oct 2014, 05:17
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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]
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17 Nov 2014, 14:03
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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. 1triplets: 6, 2triplets: 5, 3triplets: 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. 2triplet: 1, 3triplets: 2, 4triplets: 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 \(\frac{1}{6}\).
Answer A.



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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26 Aug 2016, 11:10
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] Hi, how do you throw 3 in 1 way? I have not understood these points. Will you explain these?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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27 Aug 2016, 21:54
You can throw 3 dices simultaneously to get sum as 3 in only one way  i.e. each dice should have 1 . Hence , you throw 3 in 1 way. And similarly for other numbers. NaeemHasan wrote: Hi, how do you throw 3 in 1 way? I have not understood these points. Will you explain these?
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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30 Aug 2016, 08:08
nick28 wrote: You can throw 3 dices simultaneously to get sum as 3 in only one way  i.e. each dice should have 1 . Hence , you throw 3 in 1 way. And similarly for other numbers. NaeemHasan wrote: Hi, how do you throw 3 in 1 way? I have not understood these points. Will you explain these? Now, got that. Can you describe about the triangular system as mentioned in the first reply?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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14 Sep 2016, 23:46
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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] How 15 ways to hit a 14 as per this logic?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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18 Sep 2016, 16:53
where did the 216 come from?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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30 Sep 2016, 14:48
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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] When you say that these are triangular numbers, the series is as follows: Total number  No. of ways 3 1 4 3 5 6 6 10 7 15 8 21 9 28 10 36 11 36 12 28 13 21 14 15 15 10 16 6 17 3 18 1 Adding them all, total no. of ways : 240 But for 3 throws of a dice, total no. of ways = 6x6x6 = 216 How do you account for the discrepancy?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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03 Feb 2017, 05:50
gayathri wrote: 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 If we're rolling 3 dice simultaneously, why is it that we're counting, for example, (6,1,1) as 3 ways? Is it assumed that 3 dice are distinguishable?



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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03 Feb 2017, 07:10
sharkr wrote: gayathri wrote: 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 If we're rolling 3 dice simultaneously, why is it that we're counting, for example, (6,1,1) as 3 ways? Is it assumed that 3 dice are distinguishable? Yes. Consider the die to be red, blue and green. Then 6, 1, 1 can occur in 3 ways: red  blue  green 6  1  1 1  6  1 1  1  6.
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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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01 Mar 2017, 02:00
Possible outcomes (1,1,6): 3 ways, (1,2,5): 6 ways, (1,3,4): 6 ways, (2,2,4): 3 ways, (2,3,3): 3 ways, (4,4,6): 3 ways, (4,5,5): 3 ways, (5,6,3): 6 ways, (6,6,2): 3 ways Total outcomes =36 Total outcomes = 6*6*6 = 216 Probability = 36/216 = ⅙ . Option A



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Re: What is the probability of getting a sum of 8 or 14 when [#permalink]
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09 Aug 2017, 05:18
I dont understand why there are only three ways to roll (1,1,6). There are three difference dices; therefore there is supposed to be 6 ways.



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