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

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Intern
Joined: 21 Apr 2018
Posts: 4
Re: What is the probability of getting a sum of 8 or 14 when  [#permalink]

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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 re-arrangements)?

Thanks
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Re: What is the probability of getting a sum of 8 or 14 when  [#permalink]

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21 Sep 2018, 18:09

Hello experts,

I was wondering is there any easier way to solve this problem instead of listing out each possibility?

Thank You!
Veritas Prep GMAT Instructor
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Re: What is the probability of getting a sum of 8 or 14 when  [#permalink]

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22 Sep 2018, 00:04
csaluja wrote:

Hello 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 ... l-picture/
https://www.veritasprep.com/blog/2012/1 ... e-part-ii/
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Re: What is the probability of getting a sum of 8 or 14 when  [#permalink]

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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 re-arrangements)?

Thanks

Hi Samawasthi

When 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  [#permalink]

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Updated on: 26 Sep 2018, 01:32
1
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  [#permalink]

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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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Joined: 05 Feb 2019
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What is the probability of getting a sum of 8 or 14 when  [#permalink]

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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  [#permalink]

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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   [#permalink] 14 Nov 2019, 02:59

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