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

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)   (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(52781)   (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 = :   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(3817)   (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

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