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

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

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

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

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?

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)

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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