suppose you are playing a game in which two
fair dice are rolled. you need 9 to land on
the finish by an exact count or 3 to land on
a "roll again" space. what is the probability
of landing on the finish or rolling again.
We make a chart of the sample space, all the
ways a pair of dice can fall:
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
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The sample space contains 36 outcomes or rolls.
I will color red all possible rolls in which a
sum of 3 or 9 is obtained
------------------------------------
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
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So we count 6 ways to roll the dice
such that the sum is 3 or 9, and
there are 36 ways to roll the dice.
So the desired probability is 6 ways
out of 36, or 
which reduces
to 
.
Edwin
