Probability

[gcalvin]

If you come up with something like this on an exam, you can even go right to the "brute force" method of solving it. There are 36 total combinations possible for two random six-sided die throws:

(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)

Out of those combinations, you can see that 18 of them (the second, fourth and sixth rows) satisfy the first condition (first throw even). Six combinations (the first column) satisfy the second condition (second throw 1), but three of those were in the first set also, so they need to be backed out of the total. So the probability is (18 + 6 - 3)/36 = 21/36 = 7/12.

You can also derive from this the more general rule for solving this class of problems (you wouldn't want to do much more than 36 by brute force): you take the number that satisfies the first condition, plus the number that satisfies the second condition, minus the number that satisfy both conditions (the overlaps) divided by the total number of possible combinations.

Hope that makes sense!

-Gary-