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(1) In a single toss of fair (evenly-weighted) six-sided dice, find
the probability that their sum will be at most 9.

Outcomes  (36)                  sum less than or equal to 9
--------                        ---------------------------
1 1, 1 2, 1 3, 1 4, 1 5, 1 6    6
2 1, 2 2, 2 3, 2 4, 2 5, 2 6    6
3 1, ...               , 3 6    6
4 1, ...            4 5, 4 6    5
5 1, ...       5 4, 5 5, 5 6    4
6 1, 6 2, 6 3, 6 4, 6 5, 6 6    3
                               ---
			       30

P(A) = 30/36 = 10/12 = 5/6

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(2) In a single toss of fair (evenly-weighted) six-sided dice, find
the probability that the values rolled by each die will be different
and the two dice have a sum of 6

Outcomes  (36)                  sum of 6 and values are different
--------                        --------------------------------
1 1, 1 2, 1 3, 1 4, 1 5, 1 6    1
2 1, 2 2, 2 3, 2 4, 2 5, 2 6    1
3 1, 3 2, 3 3,         , 3 6    0
4 1, 4 2, ...       4 5, 4 6    1
5 1, ...       5 4, 5 5, 5 6    1
6 1, 6 2, 6 3, 6 4, 6 5, 6 6    0
                               ---
			        4

P(A) = 4/36 = 1/9

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(3) There are 3 urns labeled X, Y, and Z.
Urn X contains 4 red balls and 3 black balls.
Urn Y contains 5 red balls and 4 black balls.
Urn Z contains 4 red balls and 4 black balls.

One ball is drawn from each of the 3 urns. What is the probability
that, of the balls drawn, 2 are red and 1 is black?

UrnX: P(red) = 4/7, P(black) = 3/7
UrnY: P(red) = 5/9, P(black) = 4/9
UrnZ: P(red) = 1/2, P(black) = 1/2

P(RRB) = 4/7 * 5/9 * 1/2 = 20/126
P(RBR) = 4/7 * 4/9 * 1/2 = 16/126
P(BRR) = 3/7 * 5/9 * 1/2 = 15/126
                           ------   
                           51/126 = 17/42