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Fall 2018 Homework 1 S&DS 241: Probability Theory with Applications Prof. Yihong WuSolutions by Ganlin Song and Yihong Wu. Blitzstein-Hwang, Chapter 1, Problem 36 Tyrion, Cersei, and ten other people are sitting at a round table, with their seating arrangement having been randomly assigned. ...

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Fall 2018 Homework 1 S&DS 241: Probability Theory with Applications Prof. Yihong Wu Solutions by Ganlin Song and Yihong Wu. 1. Blitzstein-Hwang, Chapter 1, Problem 36 36.

Tyrion, Cersei, and ten other people are sitting at a round table, with their seating arrangement having been randomly assigned. What is the probability that Tyrion and Cersei are sitting next to each other? Find this in two ways: (a) using a sample space of size 12!, where an outcome is fully detailed about the seating; (b) using a much smaller sample space, which focuses on Tyrion and Cersei. Solution : (a) Label the seats in clockwise order as 1, 2, . . . , 12, starting from some fixed seat. Give the people other than Tyrion and Cersei ID numbers 1, 2, . . . , 10. The outcome is (t, c, s1 , . . . , s10 ), where t is Tyrion’s seat assignment, c is Cersei’s, and sj is person j’s. To count the number of ways in which Tyrion and Cersei can be seated together, let Tyrion sit anywhere (12 possibilities), Cersei sit either to Tyrion’s left or to his right (2 possibilities), and let everyone else fill the remaining 10 seats in any way (10! possibilities). By the multiplication rule and the naive definition, the probability is 12 · 2 · 10! 12 · 2 · 10! 2 . = = 11 12! 12 · 11 · 10!   (b) Now let’s just consider the 122 possible seat assignments of Tyrion and Cersei, not worrying about which of these 2 seats goes to Tyrion or the details of where the other 10 people will sit. There are 12 assignments in which they sit together (without caring about order): {1, 2}, {2, 3}, . . . , {11, 12}, {12, 1}. So the probability is 12 2 12 = , 11 2

in agreement with (a).

2. Blitzstein-Hwang, Chapter 1, Problem 39 39.

Each of n balls is independently placed into one of n boxes, with all boxes equally likely. What is the probability that exactly one box is empty? Solution: In order to have exactly one empty box, there must be one empty box, one box with two balls, and n − 2 boxes with one ball (if two or more boxes each had at least two balls, then there would not be enough balls left to avoid having more than one empty box). Choose which box is empty, then which has two balls, then assign balls to the boxes with one ball, and then it is determined which balls are in the box with two balls. This gives that the probability is n(n − 1)n(n − 1)(n − 2) . . . 3 n!(n − 1) . = 2 · nn−1 nn

1

3. (Hat-checker) A hat-checker in a restaurant, having checked three hats, gets them hopelessly scrambled and returns them at random to the three guests as they leave. (a) Define an appropriate sample space Ω. List all possible outcomes. Find the cardinality |Ω|. (b) Consider the event E that exactly one guest gets his own hat back. List the outcomes that are in E. Find the cardinality |E|. Let A, B and C denote 3 guests. Let a, b and c denote 3 hat respectively. Then we have Ω = {AaBbCc, AcBbCa, AcBaCb, AaBcCb, AbBaCc, AbBcCa}. |Ω| = 6. E = {AaBcCb, AcBbCa, AbBaCc}. |E| = 3. (c) Find the probability of E . P (E) =

1 3 = . 6 2

(1)

4. A bag consists of four pairs of socks of distinct colors, say, black, white, red, and blue. Four people each pick two socks from the bag randomly without replacements. (a) What is the probability that everyone has a pair of socks of matching color? If we assume all the socks are distinct even if they have same color, and also assume every person picks the socks two times (not pick two socks at the same time), then we have |Ω| = 8!, and |E| = 4! · 24 . The probability P (E) =

1 4! · 24 . = 8! 105

(b) What is the probability that exactly one person has a pair of socks of matching color? Under the same assumptions, we have 4 choices of person who gets same color socks, 8 · 1 choices for that person to pick the socks. After he picks the socks, there are three pairs of socks (say a1,a2,b1,b2,c1,c2) and three guys (A,B,C) remaining. Each of them have to get the socks with different colors. For A, he has 6 chioces for his first pick and he only has 4 choices for his next pick since he can not get the same color socks. Let’s say he picks a1 and b1 by the symetry. Since there are four socks left (a2,b2,c1,c2), B now has 4 choices for his first pick. However, he cannot either pick (c1,c2) or (a2,b2), so no matter which socks he picks first, he only has 2 choices for his second pick. Finally, the last person C has 2 · 1 choices for his two picks. To sum up, there are (6 · 4) · (4 · 2) · (2 · 1) combinations for these there persons to pick the socks, so |E| = 4 · 8 · (6 · 4) · (4 · 2) · (2 · 1) = 212 · 3. The probability P (E) =

32 3 · 212 . = 8! 105

5. (a) Randomly permute the letters in YALE. What is the probability that the result is still YALE?

P (Y ALE) =

1 1 = . 24 4·3·2·1

2

(2)

(b) Redo the previous part for MAGA.

P (M AGA) =

1 2 = . 4·3·2·1 12

(3)

6. (de M´er´e) (a) Find the probability of getting at least a six in four tosses of a fair die. (Hint: Sometimes to find P (A), it is more convenient to first find P (Ac ).) Let A denotes at least a six in four tosses, then Ac is no six in four tosses.  4 5 ≈ 0.5177. P (A) = 1 − P (A ) = 1 − 6 c

(4)

(b) Find the probability of getting at least a double-six in twenty-four tosses of two fair dice. Let B denotes at least a double-six in twenty-four tosses, then Ac is no double-six in twenty-four tosses.   1 24 P (B) = 1 − P (B ) = 1 − 1 − 2 ≈ 0.4914. 6 c

(c) Which is one more likely? Use a calculator. At least a six in for tosses of a fair die is more likely.

3

(5)...