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

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Samawasthi

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21 Sep 2018, 10:10

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gayathri

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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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21 Sep 2018, 19:09

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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!

KarishmaB

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Updated on: 17 Oct 2022, 03:27

Originally posted by KarishmaB on 22 Sep 2018, 01:04.
Last edited by KarishmaB on 17 Oct 2022, 03:27, edited 1 time in total.

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csaluja

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!

Check out the blog posts on the link given in my signature below.

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22 Sep 2018, 01:28

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Samawasthi

gayathri

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Is there any other to solve this question ?

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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Updated on: 26 Sep 2018, 02:32

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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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

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

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22 Sep 2018, 08:17

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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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Richie Weasel

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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14 Nov 2019, 03:59

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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

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)

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25 Mar 2021, 23:46

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Can use the concept of distributing Identical items to distinct groups to determine the No. of favorable outcomes in which the 3 dice sum to either an 8 or 14

DEN = (6)^3 for each probability

(1st) Number of favorable outcomes in which the SUM of the 3 dice = 8

Call the 3 die - A , B, and C

A + B + C = 8

Each die must show at least a 1 integer on its face. So we can begin by distributing “one identical chocolate” to each of the die

a + b + c = (8 - 3) = 5

No of ways 3 thrown dice can sum to 8 is equivalent to the number of ways we can distribute 5 identical chocolates among 3 distinct children

Also note, we do not have any issue here if any one die receives all of the identical “chocolates”.

6, 1, 1 ———> is a possibility

7! / (5! 2!) = 21 favorable outcomes

OR

(2nd) number of ways 3 dice can sum to = 14

A + B + C = 14

Again, we’ll begin by giving each die “one identical chocolate/one” because each die must show at least a 1 integer

a + b + c = (14 - 3) = 11

However, now there is a problem. Each lower case variable already contains ONE 1

We can not distribute the value of 11 to each of the distinct die with no constraints because we will count the possibilities such as:

12 , 1 , 1

And other such possibilities in which any given die can show a number greater than > 6

a + b + c = 11

For each variable, we will substitute the following:

a = 5 - x

b = 5 - y

c = 5 - z

This way the amount distributed will not exceed 6 to any one die

(5 - x) + (5 - y) + (5 - z) = 11

15 - (x + y + z) = 11

x + y + z = 4

Now, when we distribute the identical items, it does not matter if one die receives all 4 or one die receives none (because we already gave one to each die)

For instance if:

x = 0
y = 4
z = 0

a = 5 - 0 = 5 ——-> since Die A already received one, A will show a 6

b = 5 - 4 = 1 ———> since Die B already received one B will show a 2

c = 5 -0 = 5 ———> Die C will show a 6

6 , 2 , 6 ———> sums to 14

The number of ways to distribute 4 identical items to 3 distinct groups:

6! / (4! 2!) = 15 ways

21 + 15 = 36 favorable ways

Probability = 36 / (6)^3 =

1/6

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