Exercise Lover

March 24, 2007

Exercises C.2-10

Filed under: C.2 — yuhanlyu @ 11:57 am

Three prisoners problem.
Six cases, which upper case means which one will be executed, and the name in bracket is guard’s answer.
Case 1: X Y z [Y]
Case 2: X Y z [Y]
Case 3: X y Z [Z]
Case 4: X y Z [Z]
Case 5: x Y Z [Y]
Case 6: x Y Z [Z]

In the original, the probability of X is 2/3. After guard telled Y is executed, the probability is the same.

Exercises C.2-9

Filed under: C.2 — yuhanlyu @ 11:40 am

Monty Hall Problem.
Not change = 1/3
Change = 2/3

Exercises C.2-8

Filed under: C.2 — yuhanlyu @ 11:37 am

Toss two biased coins. Let A be the first coin’s outcome, B be the second coin’s outcome. Since coins are biased, A and B are dependent. Therefore, let C be the coin baised toward head. If C is given, then A and B are independent.

Exercises C.2-7

Filed under: C.2 — yuhanlyu @ 11:31 am

Solution wanted!

Exercises C.2-6

Filed under: C.2 — yuhanlyu @ 11:23 am

Principle of deferred decisions.
Proof by induction.
Base case: \Pr(A_1 \cap A_2) = \Pr(A_1) \Pr(A_2 | A_1).
Induction case: Assume that \Pr(\cap^n_{i=1}A_i) = \Pr(A_1) + \Pr(A_2 | A_1) + \dots for all n < k. Then \Pr(\cap^k_{i=1}A_i) = \Pr(A_1) + \Pr(A_2 | A_1) + \dots + \Pr(A_k|\cap^{k-1}_{i=1} A_i).

Exercises C.2-5

Filed under: C.2 — yuhanlyu @ 11:07 am

\Pr(A|B) + \Pr( \bar{A} | B) = \Pr(A \cap B)/\Pr(B) + \Pr(\bar{A} \cap B) / \Pr(B) = 1

Exercises C.2-4

Filed under: C.2 — yuhanlyu @ 10:26 am

Solution wanted!

Exercises C.2-3

Filed under: C.2 — yuhanlyu @ 10:25 am

\frac{{10 \choose 3}}{{10 \choose 3}3!} = 1/6

Exercises C.2-2

Filed under: C.2 — yuhanlyu @ 10:05 am

The only possibility is R: Head G: Tail, Tail. The probability is 1/8.

Exercises C.2-1

Filed under: C.2 — yuhanlyu @ 9:52 am

Union bound.
Proof by induction.
Base case: \Pr(A_1 \cup A_2) = \Pr(A_1) + \Pr(A_2) - \Pr(A_1 \cap A_2) \leq \Pr(A_1) + \Pr(A_2).
Induction step: Assume that \Pr( \cup^n_{i=1} A_i) \leq \sum^n_{i=1} \Pr(A_i) for all n < k.
Then \Pr( \cup^k_{i=1} A_i) = \Pr( \cup^{k-1}_{i=1} A_i \cup A_k) = \Pr( \cup^{k-1}_{i=1} A_i) + \Pr(A_k) - \Pr(\cup^{k-1}_{i=1} A_i \cap A_k) \leq \sum^k_{i=1} \Pr(A_i).

Blog at WordPress.com.