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.

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One response to this post.

  1. Posted by Man on June 30, 2017 at 6:06 pm

    How do you get (c) of Exercise 1.16? Thanks.

    Reply

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