27) Let X = face value of first die
Y = face value of second die.
There are 6 * 6 = 36 possible outcomes as follows:
Y|X | 1 | 2 | 3 | 4 | 5 | 6 |
1 | (1,1) | (2,1) | (3,1) | (4,1) | (5,1) | (6,1) |
2 | (1,2) | (2,2) | (3,2) | (4,2) | (5,2) | (6,2) |
3 | (1,3) | (2,3) | (3,3) | (4,3) | (5,3) | (6,3) |
4 | (1,4) | (2,4) | (3,4) | (4,4) | (5,4) | (6,4) |
5 | (1,5) | (2,5) | (3,5) | (4,5) | (5,5) | (6,5) |
6 | (1,6) | (2,6) | (3,6) | (4,6) | (5,6) | (6,6) |
28) There are 11 possible outcomes of X + Y as 2, 3 , 4, 5, 6, 7, 8, 9, 10, 11, and 12
29) P( X + Y = 6) = 5/36 = 0.1389
Because total possible outcomes = 5 as {(1,5 ); ( 2, 4); ( 3, 3); (4, 2); (5,1)}
And total outcomes = 36
30) P( X + Y < 6) = P(X + Y = 2) + P(X + Y = 3) + P(X + Y = 4) + P(X + Y = 5)
= (1/36)+(2/36)+(3/36)+(4/36) = 10/36 = 2.2778
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