euro wrote:(Q) 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

The official answer is [spoiler](a) [/spoiler]

Can somebody explain?

kmittal82 wrote:Wow, tough one...

Lets look at it case by case

#1) To get 8

8 = :

1+1+6
2+2+4
3+3+2
1+2+5
1+3+4


Since order won't matter, any combination of those 3 numbers will give you a sum of 8. Each combination can further have 6 arrangements

Stuart Kovinsky wrote:

kmittal82 wrote:Wow, tough one...

Lets look at it case by case

#1) To get 8

8 = :

1+1+6
2+2+4
3+3+2
1+2+5
1+3+4


Since order won't matter, any combination of those 3 numbers will give you a sum of 8. Each combination can further have 6 arrangements

thanks for the detail expanation but i thought the dices are different , of course it have the same numbers but its from different dice

The bolded statement is your error.

For three different digits, such as 1+3+4, there are indeed 6 different arrangements. However, for the cases that have only 2 distinct digits, there are only 3 different arrangements.

So:

1/1/6 - 3 arrangements
2/2/4 - 3 arrangements
3/3/2 - 3 arrangements
1/2/5 - 6 arrangements
1/3/4 - 6 arrangements

Total - 21 arrangements

Similarly for 14:

6/6/2 - 3 arrangements
6/4/4 - 3 arrangements
5/5/4 - 3 arrangements
6/5/3 - 6 arrangements

Total - 15 arrangements

Total possibilities is 6^3 = 216

Prob = # of desired outcomes / total # of possibilities

Prob = (21 + 15)/216 = 36/216 = 1/6