## Probability and Computing: Chapter 1 Exercises (Cont. 3)

Exercise 1.14:

Let $A$ be the event that we are equally talented,  $B$ be the event that I am slightly better, $C$ be the event that he is slightly better, $D$ be the event that the game ends up in one win for me and three wins for him.

We have:

$Pr(A) = Pr(B) = Pr(C) = \frac{1}{3}$

$Pr(D) = Pr(D\mid A)Pr(A) + Pr(D\mid B)Pr(B) + Pr(D\mid C)Pr(C)$

$\frac{1}{3}\cdot (\frac{1}{2})^{4} + \frac{1}{3}\cdot 0.6 \cdot (0.4)^{3} + \frac{1}{3} \cdot 0.4 \cdot (0.6)^{3} = 0.0624$

Apply Bayes’ Law:

$Pr(C\mid D) = \frac{Pr(D\mid C)Pr(C)}{Pr(D)}=\frac{0.0864\cdot \frac{1}{3}}{0.0624} = 0.4615$

Therefore in my posterior model, I should believe that my opponent is slightly better than I am with probability $0.4615$.

Exercise 1.15:

Let A be the event that their sum is divisible by 6.

The principle of deferred decision yields $Pr(A) = \frac{1}{6}$.

Exercise 1.16:

Let $E_{i,j}$ be the event that there are exactly $i$ of three dice have the same number on the $j$-th roll, B be the event that the player wins the game.

(a) $Pr(E_{3,1})= \frac{1}{6}\cdot \frac{1}{6} = \frac{1}{36}$

(b) $Pr(E_{2,1}) = 3\cdot (\frac{1}{6} \cdot \frac{5}{6}) = \frac{15}{36}$

(c) $Pr(B\mid E_{2,1}) = \frac{1}{6} + \frac{5}{6}\cdot\frac{1}{6} = \frac{11}{36}$

(d)

In order to calculate $Pr(B)$ we calculate two probabilities $Pr(B\mid E_{1,1})$ and $Pr(B\mid E_{3,1})$

$Pr(B\mid E_{3,1}) = 1$

$Pr(B \mid E_{1,1}) = \frac{1}{36} + \frac{15}{36}\cdot \frac{1}{6} + \frac{20}{36}\cdot \frac{1}{36} = 0.113$

$Pr(B)$

$=Pr(B\mid E_{1,1})Pr(E_{1,1}) + Pr(B \mid E_{2,1})Pr(E_{2,1}) + Pr(B \mid E_{3,1})Pr(E_{3,1})$

$=0.113\cdot \frac{20}{36} + \frac{11}{36}\cdot \frac{15}{36} + \frac{1}{36}$

$= 0.218$

Exercise 1.20: Show that if $E_{1},E_{2},\ldots ,E_{n}$ are mutually independent, then so are $\bar{E_{1}},\bar{E_{2}},\ldots ,\bar{E_{n}}$.

We only prove for $n =2$.

$Pr(\bar{E_{1}})Pr(\bar{E_{2}}) = (1-Pr(E_{1}))(1-Pr(E_{2}))$

$=1 - (Pr(E_{1} )+ Pr(E_{2}) - Pr(E_{1})Pr(E_{2}))$

$= 1-(Pr(E_{1})+Pr(E_{2})-Pr(E_{1}\cap E_{2}))$

$= 1-Pr(E_{1} \cup E_{2})$

$= Pr(\overline{E_{1} \cup E_{2}}) = Pr(\bar{E_{1}} \cap \bar{E_{2}})$

Since $Pr(\bar{E_{1}})Pr(\bar{E_{2}}) = Pr(\bar{E_{1}} \cap \bar{E_{2}})$, $\bar{E_{1}}$ and $\bar{E_{2}}$ are independent.