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

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

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New post 21 Sep 2018, 10:10
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)?

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

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New post 21 Sep 2018, 19:09
chetan2u Bunuel VeritasKarishma

Hello 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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New post 22 Sep 2018, 01:04
csaluja wrote:
chetan2u Bunuel VeritasKarishma

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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New post 22 Sep 2018, 01:28
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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New post Updated on: 26 Sep 2018, 02:32
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, 03:05.
Last edited by dave13 on 26 Sep 2018, 02:32, 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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New post 22 Sep 2018, 08:17
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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Re: What is the probability of getting a sum of 8 or 14 when &nbs [#permalink] 22 Sep 2018, 08:17

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