Problem #1 #2 #3 #4 Extra Total Points /3 /13 /7 /10 /4 /33

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1 STAT/MATH 394 A - Autumn Quarter Midterm - October 2, 206 Name: Student ID Number: Problem # #2 #3 #4 Extra Total Points /3 /3 /7 /0 /4 /33 Read directions carefully and show all your work. Particularly, specify the events you are considering. Partial credit will be assigned based upon the correctness, completeness and clarity of your answers. Correct answers without proper justification will not receive full credit. The exam is closed book, closed notes. Calculators are permitted but not needed to answer questions. Problem.[3 points] Complete the three axioms of Probability. Definition. Let (Ω, A be a measurable space of events. A probability measure is a real-valued function mapping P : A R satisfying : (a [ point] for any event E A, P(E 0 (b [ point] P(Ω = (c [ point] for any countably infinite sequence (E i i of mutually exclusive events (i.e. E i E j = if i j, ( P E i = P(E i i= i=

2 Problem 2.[3 points] Consider an experiment that consists of determining the type of job (either blue collar or white collar and the political affiliation (either Republican, Democratic or Independent of 5 members of an adult soccer team. (a [2 points] Give a formal description of Ω. Answer. We denote B and W for blue collar and white collar respectively. The political affiliations, Republican, Democratic or Independent are respectively denoted by R, D and I. The sample space can be represented by the following set: Ω = {((j, p,..., (j 5, p 5 j k {B, W }, p k {R, D, I}} = ({B, W } {R, D, I} 5 How many outcomes are (b [2 points] in the sample space Ω? Answer. By the rule of product, Ω = ( (c [2 points] in the event that none of the team members considers himself or herself an Independent? Answer. Let E be the event that none of the team members considers himself or herself an Independent. If E occurs, then all the members are either Republican or Democratic. Hence, E = ({B, W } {R, D} 5 and E = ( (d [3 points] in the event that at least one of the team members is a blue-collar worker? Answer. Let F be the event that at least one of the team members is a blue-collar worker. It is easier to consider the event F c, which is that all the team members are white-collar workers, i.e. F c = ({W } {R, D, I} 5. Therefore, F = Ω F c = 6 5 ( 3 5. (e [4 points] Now assume that in the team there are 9 blue-collar workers and 7 are politically Independent and 3 are both blue collar and Independent. Find the probability that a randomly selected member of the team is neither blue collar nor Independent. Answer. P(B c I c = P((B I c = P(B I = P(B P(I + P(B I =

3 Problem 3.[7 points] Assume that n people (n N, n 2 are in a room. (a [2 points] According to the pigeonhole principle, what is the smallest value of n such that we are sure that at least two people in the room celebrate their birthday in the same month? No justifications required for this question. Answer. 3 (b [5 points] Now assume that 2 n 2. Find the probability that among the n people, there is no pair of them sharing the same birth month. Answer. First, the sample space Ω consists of all possible birth months for each of the n people in the room. Therefore Ω = 2 n. Next, you can think of the problem using the following analogy: people are items and months are boxes. The event E that no one shares the same birth month as anyone else is equivalent to putting n people into distinct boxes. Therefore, you have 2 possible choices of months for the first person, for the second, and so on, and lastly 2 n + choices for the n-th person. By the rule of product, E = 2... (2 n + = 2!/(2 n!. Hence, the requested probability is given by P(E = 2!/(2 n! 2 n

4 Problem 4.[0 points] During a film festival, six awards are to be given out among 40 films. A film can receive multiple awards. Assume that each film is assigned a number between and 40. A possible outcome of the experiment is given by a vector (a, a 2, a 3, a 4, a 5, a 6 where a i {, 2,..., 40} is the number of the film that wins the i-th award. For instance, the vector (4, 3, 27, 4, 3, 4 represents the outcome that film number 4 wins the first, the fourth and the sixth awards, film number 3 wins the second and the fifth awards and film number 27 wins the third award. (a [2 points] What is the size of the sample space Ω? Answer. Clearly, Ω = 40 6 If we assume that each film is equally likely to receive any of the six awards, what is the probability that (b [4 points] one film receives exactly four awards? Answer. Let E be the event that one film receives exactly four awards. E can be decomposed into 3 sub-events: first, choose the film that receives the 4 awards, then assign 4 awards among 6 to the chosen film and finally the last 2 awards can be given out among any of the 39 other films. Therefore, E = ( 40 4( 39 ( 39. P(E = 40( (c [4 points] no film receives more than four awards? Answer. Let F be the event that no film receives more than four awards. Consider F c the complement of F, which is that one film receives at least 5 awards. F c can be split into two mutually exclusive events: the event that one film receives exactly 5 awards and the event that one film receives exactly all of the 6 awards. Using the same reasoning as in the previous question, we have that F c = ( 40 5( 39 + ( P(F c = 40(

5 Extra-credit problem. [4 points] Let E and F be two events on a given sample space Ω. Suppose now that we have the following two assumptions. E F ; P(E = P(F E c. Prove that P(E = P(F /2. Answer. We will apply the law of total probability on F with the partition {E, E c }. P(F = P(F E + P(F E c = P(E + P(E = 2P(E. Note that F E = E since E F.

STAT:5100 (22S:193) Statistical Inference I

STAT:5100 (22S:193) Statistical Inference I Week 3 Luke Tierney University of Iowa Fall 2015 Luke Tierney (U Iowa) STAT:5100 (22S:193) Statistical Inference I Fall 2015 1 Recap Matching problem Generalized