for sum of 8 we will haveeuro 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?
The bolded statement is your error.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:thanks for the detail expanation but i thought the dices are different , of course it have the same numbers but its from different dicekmittal82 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
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
