SOLUTION: a standard six faced die and a standard four-faced die are both rolled once the cube’s faces are numbered 1-6 and the tetrahedron ‘s faces are numbered 1-4 what is the probability

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Question 269124: a standard six faced die and a standard four-faced die are both rolled once the cube’s faces are numbered 1-6 and the tetrahedron ‘s faces are numbered 1-4 what is the probability that the sum of the downward – facing faces is grater than 6 ?
Found 2 solutions by AnlytcPhil, stanbon:
Answer by AnlytcPhil(1807)   (Show Source): You can put this solution on YOUR website!
a standard six faced die and a standard four-faced die are both rolled once the cube’s faces are numbered 1-6 and the tetrahedron ‘s faces are numbered 1-4 what is the probability that the sum of the downward – facing faces is greater than 6 ?

Here is the sample space, where, for example (4,3) stands for
getting a downward facing 4 on the six-faced die and a downward 
facing 3 on the four-faced die:  

{ (1,1)  (1,2)  (1,3)  (1,4)

  (2,1)  (2,2)  (2,3)  (2,4)

  (3,1)  (3,2)  (3,3)  (3,4)

  (4,1)  (4,2)  (4,3)  (4,4)

  (5,1)  (5,2)  (5,3)  (5,4)

  (6,1)  (6,2)  (6,3)  (6,4)  }

There are 24 outcomes or simple events in the sample space.
I'll color the ones red which have a sum greater than 6

{ (1,1)  (1,2)  (1,3)  (1,4)

  (2,1)  (2,2)  (2,3)  (2,4)

  (3,1)  (3,2)  (3,3)  (3,4)

  (4,1)  (4,2)  (4,3)  (4,4)

  (5,1)  (5,2)  (5,3)  (5,4)

  (6,1)  (6,2)  (6,3)  (6,4)  }

Count the red ones and you'll get that 10 out of the 24 
have a sum greater than 6. So the probability you want is

 which reduces to 

Edwin
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
a standard six faced die and a standard four-faced die are both rolled once the cube’s faces are numbered 1-6 and the tetrahedron ‘s faces are numbered 1-4 what is the probability that the sum of the downward – facing faces is greater than 6 ?
-------------------
cube values: 1,2,3,4,5,6
tetra values: 1,2,3,4
---------------------------
Sum greater than 6
6 with 1,2,3,4
5 with 2,3,4
4 with 3,4
3 with 4
-----
There are 10 ways to succeed
----
There are 4*6 possible outcomes
----
P(greater than 6) = 10/24 = 5/12
==========================================
Cheers,
Stan H.
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