Rolling two dice
A standard six-faced die and a standard four-faced die are both rolled once. The cube's faces are numbered , and the tetrahedron's faces are numbered . What is the probability that the sum of the downward-facing faces is greater than ? Give your answer correct to one decimal place.
The answer is 0.4.
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1 solution
Let's denote outcomes by ordered pairs in which the first coordinate denotes the downward-facing face of the six-faced die and the second coordinate denotes the downward facing face of the four-faced die. Here is the sample space, there are a total of 2 4 outcomes with 1 0 outcomes having a sum greater than 6 (colored red).
( 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 )
Since all outcomes are equally likely, the probability of achieving a sum greater than 6 is
P = 2 4 1 0 = 1 2 5 ≈ 0 . 4