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Re: What is the probability of getting a sum of 8 or 14 when
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21 Sep 2018, 09:10
gayathri wrote: 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. Hi Can you please explain why there are 6 options as opposed to 3 options(3 rearrangements)? Thanks



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Re: What is the probability of getting a sum of 8 or 14 when
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21 Sep 2018, 18:09
chetan2u Bunuel VeritasKarishmaHello experts, I was wondering is there any easier way to solve this problem instead of listing out each possibility? Thank You!



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Re: What is the probability of getting a sum of 8 or 14 when
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22 Sep 2018, 00:04
csaluja wrote: chetan2u Bunuel VeritasKarishmaHello experts, I was wondering is there any easier way to solve this problem instead of listing out each possibility? Thank You! Yes, I have discussed in detail how to find every possible sum on a roll of 3 dice in these two posts: https://www.veritasprep.com/blog/2012/1 ... lpicture/https://www.veritasprep.com/blog/2012/1 ... epartii/
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Re: What is the probability of getting a sum of 8 or 14 when
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22 Sep 2018, 00:28
Samawasthi wrote: gayathri wrote: 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. Can you please explain why there are 6 options as opposed to 3 options(3 rearrangements)? Thanks Hi SamawasthiWhen we are taking the possibility of suppose (6,6,2), the number of ways to arrange these 3 numbers would be 3!/2! = 3 ways (Divided by 2! because 6 is repeated twice) (6,6,2) can be written as (2,6,6), (6,6,2) and (6,2,6)  3 options/ways. When we are taking possibility of suppose (6,5,3), the number of ways to arrange these 3 different numbers would be 3! = 6 ways (We have no similar numbers) (6,5,3) can be written as (6,5,3), (5,6,3), (6,3,5), (3,5,6), (3,6,5) and (5,3,6)  6 options/ways
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What is the probability of getting a sum of 8 or 14 when
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Updated on: 26 Sep 2018, 01:32
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 hey pushpitkc , can you please let me know if my reasoning and approach is correct ? if not please identify the flaws:) thank you and have a great weekend Number of Ways to get 8 (not to get confused and or not to miss some number, start listing number in increasing order ) 116 125 134 143 152 161 now i see that there are 6 ways, but since each combination has 3 numbers, i can reranrange each combination in 3 ways Hence 6 * 3 = 18 to calculate number of ways to get sum 14, i didnt have to enumerate favorite outcomes, because logically it must be the same number of ways as is the case with 8 Hence 6 * 3 = 18 So total number of favoroble outcomes is 18+18 = 36 Total Number of Ways: so we have three dices choosing any number from first dice: is 1/6 choosing any number from second dice: is 1/6 choosing any number from third dice is 1/6 hence total number of way \(\frac{1}{6} * \frac{1}{6} * \frac{1}{6}\) = \(\frac{1}{216}\) probability of getting a sum of 8 or 14 when rolling 3 dice simultaneously is \(\frac{36}{216}\) i.e =\(\frac{1}{6}\) hey pushpitkc are you there ? need your help clearing confusions I have a doubt regarding total number of ways I wrote above, in one of the posts I saw that total number of ways is 6*6*6 ok dice has 6 dfferent numbers, and we have three dices: when we say we can get any number in 6 ways, why we write 6*6*6 and not 6! *6!*6! ? can you explain the difference? and why multiplying \(\frac{1}{6}\) by itself three times is incorrect ?Hello niks18 may be you can help me to clear my confusion. see the highlighted part above thank you
Originally posted by dave13 on 22 Sep 2018, 02:05.
Last edited by dave13 on 26 Sep 2018, 01:32, edited 7 times in total.



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Re: What is the probability of getting a sum of 8 or 14 when
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22 Sep 2018, 07:17
i did it manually. But if you look at pattern it is easy so three dices are rolled. We need the sum to be 8 or 14 start systematically Sum = 8 161 251 341 431 521 611 152 242 332 422 512 143 233 323 413 134 224 314 125 215 116 Total = 21 nos Sum =14 266 356 365 446 455 464 536 545 554 563 626 635 664 653 662 Total = 15 Nos hence total cases = 21+15 = 36
Hence the probability = 36/216 = 1/6 Answer



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What is the probability of getting a sum of 8 or 14 when
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23 Apr 2019, 07:23
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 did you get 15 ways to get a 14?



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What is the probability of getting a sum of 8 or 14 when
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14 Nov 2019, 02:59
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 TOTAL OUTCOMES: \(6^3=216\) POSSIBLE OUTCOMES: for sum of 8 and 14 \(=36\) [116=8]: arrangements(3!/2!)=3 [125]: arrangements(3!)=6 [134]: 6 [224]: 3 [233]: 3 Possible outcomes for sum of 8: 3+6+6+3+3=12+9=21 [662=14]: 3 [653]: 6 [644]: 3 [554]: 3 Possible outcomes for sum of 14: 3+6+3+3=6+9=15 PROBABILITY: (possible/total)=36/216=12/72=4/24=1/6 Ans (A)




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