Sample space = {p, r, o, b, a, b, i, l, I, t, y} = 11 possible outcomes 1sr event: drawing an I ( there are 2 I); P(1st I) = 2/11 2nd event drawing also an i: This is a conditional probability, since one I has already been selected the remaining number of I is now 1, but also the sample space from previously 11 outcome has now 10 outcomes (one letter selected and not replaced) 2nd event : P(also one I) = 1/10 P(selecting one I AND another I) is 2/11 x 1/10