Probability Theory and Random Processes2009-09-10Ben Blum (bblum)1. We know that P(A|B) = P(AnB)/P(B). Then, P(AnB|B) = P((AnB)nB)/P(B) = P(AnB)/P(B) = P(A|B) qed2. 1, since that is the entire sample space.3. 1/6 + 5/6*1/6 = 6/36+5/36 = 11/364. sample space: 6+5+4 possibilities = 15 possibilities; first die 6 in three sums: 3 ways, so 3/15 => 1/55. 5 white vs 3 white: 3/8 landed tails.6. a) 1/2 b) 1/37. sample space given colourblind: 5+25 = 30; P(male|colourblind) = 5/30 = 1/68. P(one or the other) = P(car u home) - P(car n home) = 60+30-20-20 = 50%9. A_i are not independent, because (wlog) if A_1 and A_2, then result_2 = result_3 (by A_1) and result_1 = result_3 (by A_2); therefore, result_2 = result_1 by substitution and therefore A_3 as well. They are, however, pairwise independent, because for A_i, the value of result_i does not matter, and A_j which depends on result_i can have it be whatever it wants.10. In order for it to remain independent, the proportions must stay the same between all the fields, so 4/6 = 6/x => 9 sophomore girls.