SOLUTION: An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Using the sample space provided below and assuming each simple event is as likely as

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Question 1197526: An experiment consists of rolling two fair dice and adding the dots on the two sides facing up. Using the sample space provided below and assuming each simple event is as likely as any​ other, find the probability that the sum of the dots is not 3 or 9 .
Found 3 solutions by ewatrrr, ikleyn, math_tutor2020:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!

Hi
An experiment consists of rolling two fair dice and adding the dots on the two sides facing up
P(sum is not 3 OR 9) = 30/36 = 5/6
 -----------------
___1__2__3__4__5__6
1|_2__3__4__5__6__7
2|_3__4__5__6__7__8
3|_4__5__6__7__8__9
4|_5__6__7__8__9__10
5|_6__7__8__9__10_11
6|_7__8__9_10__11_12 
 
Wish You the Best in your Studies.


Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.
An experiment consists of rolling two fair dice and adding the dots on the two sides facing up.
Using the sample space provided below and assuming each simple event is as likely as any​ other,
find the probability that the sum of the dots is not 3 or 9.
~~~~~~~~~~~~~~ 

   +---------------------------------------------------------------------------+
   |                    I read your question as                                |
   | "find the probability that the sum of the dots is NEITHER 3 NOR 9",       |
   |              as it is normally accepted in English.                       |
   +---------------------------------------------------------------------------+


The answer is  30%2F36 = 5%2F6 :   from the total  36  cases,  we should exclude  2 + 4 = 6 cases,

2  cases with the sum of  3,  and  4  cases with the sum of  9.


Solved  with full explanations.



Answer by math_tutor2020(3816) About Me  (Show Source):
You can put this solution on YOUR website!

Here are the ways to add to 3
1+2 = 3
2+1 = 3
and here are the ways to add to 9
3+6 = 9
4+5 = 9
5+4 = 9
6+3 = 9

There are 2+4 = 6 cases in which we get either 3 or 9
There are 6*6 = 36 dice rolls total
That leaves 36-6 = 30 cases in which we do not get a sum of 3, nor do we get a sum of 9 either.

This leads to 30/36 = (5*6)/(6*6) = 5/6

Answer: 5/6