1. We know that $P(A)\ge 0$, so we assume the opposite, i.e. $P(X\ge \mu)=0$ so $X<\mu$ with 100% certainity, we get $\mu=\sum_{x}xp(x)=\sum_{x< \mu}xp(x)$ and $\displaystyle\sum_{x<\mu}xp(x)<\sum_{x<\mu}\mu p(x)=\mu\sum_{x<\mu}p(x)=\mu P(X<\mu)=\mu$, a contradiction $\mu<\mu$. Similarly for the opposite case [Reference: Probability and Computing: Randomized Algorithms and Probabilistic Analysis by Michael Mitzenmacher, Eli Upfal (Lemma 6.2)]
  2. TODO [Reference: Probability and Computing: Randomized Algorithms and Probabilistic Analysis by Michael Mitzenmacher, Eli Upfal (Theorem 6.3)]
  3. TODO (SAT not taught yet) [Reference: Probability and Computing: Randomized Algorithms and Probabilistic Analysis by Michael Mitzenmacher, Eli Upfal (Theorem 6.4)]
  4. \(E[M]=\underbrace{\frac 26}_{1,2}\cdot 2+\underbrace{\frac 16}_{5}\cdot 10=2.\bar 3\) Unfair.
  5. \(E[M]=\underbrace{\frac 1{13}}_{\text K}\cdot 4+\underbrace{\frac 2{13}}_{5,10}\cdot 3+\underbrace{\frac 3{13}}_{3,6,9}\cdot 2\approx 1.230\) Unfair.
  6. \(E[M]=\underbrace{\left(\frac{2\cdot4}{36}\right)}_{(1,3),(2,4),(3,5),(4,6)}\times 2+\underbrace{\left(\frac{2\cdot1}{36}\right)}_{(1,6)}\times 5= 0.7\bar2\) Unfair.