HW1

Author: Freeman Chen §

Data: 2023-07-07 §

Class: MSDS 504 §

---

(a). The outcome space of this experiment is the the sum of top faces that recorded which are {2,3,4,5,6,7,8,9,10,11,12}

(b). The sets that outcome will be 5 : {(dice1 = 1, dice2 = 4),(dice1 = 4, dice2 = 1),(dice1 = 2, dice2 = 3),(dice1 = 3, dice2 = 2)}

(c) The sets that outcome will be divisible by 5 will be the sum that equal to 5 and 10. since we are only counting (5,5) once so the number of favorable outcome will be 7. P(outcome will be divisible by 5 ) =

---

From the Theorem we know that for is a probability measure function, it must satisfy: - , - - if are mutually exclusive (a) Prove Q is a probability measure on S: - Since is a probability measure of sample space , satisfy the Theorem above. Given that , with , . would also satisfy this condition.. - the CDF of the interval is squared, it still continuous and will still get to 1. Because there will exist a value such that , and when we square that we should still get 1. - Since are mutually exclusive. Therefore, let and . or it could rewrite as Contradiction.Therefore, Q is not a probability measure on S. (b) Prove R is a probability measure on S: - Scine ,Then , thus - Since and is a probability measure, so there exist a value such that , , thus R is not a probability measure for S

---

(a). (b). (c).

---

From the theorem5, gives that Prove: Give (Apply theorem5) (Apply theorem5)

---

(a) Four games {AAAA, NNNN} 2 orders (b)Six games: base on the rule, the last game is a must win for the team who win in 6 so the combination function is thus, there are 20 orders to win

---

6. The percent of the unemployed are woman: 60% of the unemployed are women.

---

7. Total card in a deck: 52, 4 card is ace, the chance of second card is an - first case:
   • first case: draw an ace from the first draw, then the second draw is also an Ace:
   • second case: the first card is not ace, the second one is ace:

---

8. The probability of the boy choose box 1 & box 2 is , the probability of box 1 for blue with picking box1 , the probability of box 2 for blue with picking box 2 , and the probability for the boy to pick the blue ball is

---

9.(a) given that thus . (b)
(c)

---

10.(a)

(b) at least on match = Apply # Inclusion–exclusion principle

-> =

(c) =

---

(a) Let A and B are mutually exclusive, then , for and be independent, and must satisfy thus, mutually exclusive events can’t be independent,unless the probability of one event is zero (b) Let and be independent, then . as (a), the independent events can’t be mutually exclusive unless the probability of one event is zero (c) Let , then . If and are independent, then .if P(A) = 0 or P(B) =1, then the equality hold.