Question 1118623
a)  Pr{E1} = 6/36 = <b>1/6</b>  (there are 6 ways to roll a 7 out of 36 possible outcomes)<br>
b)  Pr{E2} = 3/6  = <b>1/2</b>  (there are 3 ways to roll an odd number out of 6 possible outcomes)<br>
c)  Pr{E1 &#8745; E2} = 3/36 = <b>1/12</b>  (3 out of 36 possible outcomes fall in the intersection, or, one can multiply the results of (a) and (b))<br>
d)  Pr{E1 U E2 } = 21/36 = <b>7/12</b>   (this is the set of elements that comprise E1, E2, or both E1 & E2.  We can count them: E1: Die1={1, 3, 5} has 18 outcomes, and E2 {sum of Die1 and Die2 = 7} has 6 outcomes, but we've over counted by 3 outcomes (where Die1 is odd and Die1+Die2 = 7).  One can also compute (a) + (b) - (c) = 6/36+18/36 - 3/36 = 21/36 = 7/12.