2. Note, Pr[C u D] = Pr[C] + Pr[D] - Pr [C ^ D]
a) Pr[C | D] = Pr[C ^ D] / Pr[D] = 3/4,
b) Pr[D | C] = Pr[D ^ C] / Pr[C] = 3/8.
6. Note, if independent, then Pr[A ^ B] = Pr[A]
Pr[B]. Therefore, Not Independent.
10. a) 0.50, b) 0.40.
14. a) 0.30, b) 0.40.
18. Pr[one is 3 | sum is 5] = Pr[one is 3 ^ sum is 5]
/ Pr[sum is 5] = (2/36) / (4/36) = 1/2.
22. 4/5.
26. Pr[D] = Pr[AAA]Pr[D | AAA] + Pr[AA]Pr[D | AA] +
Pr[A]Pr[D | A] = 0.0075 + 0.03 + 0.11 = 0.1475.
30. (Number of ways to assign C) x (Number of ways to
assign PT a number that will not play C until round 3) x (number of
ways to assign other 6 teams) / (number of ways to assign 8 teams a
number) = (8) x (4) x (6!) / (8!) = 4/7.
