Homework 3 (due Tuesday, October 13)


 Bartholomew Harvey
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1 Homework (due Tuesday, October 1 Problem 1. Consider an experiment that consists of determining the type of job either bluecollar or whitecollar and the political affiliation Republican, Democratic, or Independent of the 1 members of an adult soccer team. How many outcomes are (a in the sample space; Each of the players has 2 possible job types and possible political affiliations, So a total of 6 possible choices for each of the 1 players. Thus a total of 6 1 possible outcomes for the soccer teams job and political composition. (b in the event that at least one of the team members is a bluecollar worker; If one wants to count the possibilities where at least one is a blue collar worker we count the ones in which no one is a blue collar worker from the total 6 1. If no one is to be a bluecollar worker then for each player there is only one choice of job type and three choices for political affiliation. Thus 1 total where no one is a bluecollar worker. Thus the number where there is at least one blue collar worker is (c in the event that none of the team members considers himself or herself as Independent? If no one considers themselves independent then there are only 2 possible choices for political choices and 2 job types. Thus for each player a possible of 4 different combos of job and political affiliations. Thus, there are 4 1 different outcomes of this type. Problem 2. probability of being dealt If it is assumed that all ( 2 poker hands are equally likely, what is the (a a full house? (A hand is said to be a full house if the hand has the values a, a, b, b, b. ( 2 The total number of hands one could be dealt without restriction is. This is the size of our sample space. Now choose from the 1 cards types the numbers for ( 1 the a s and the numbers for the b s from the remaining 12. Now for the a s 1 choose the 2 suits and suits for the b s. Thus the probability of getting 2 such a hand is: ( 2 =
2 (b two pairs? (This occurs when the cards have values a, a, b, b, c, where a, b, c are all distinct. To count the outcomes that look like this we choose the value that are to appear ( 1 ( in our hand. Then from those choose the 2 that are to be the pairs =. 2 Then from each of the 2 values that are to be the pairs you pick the 2 in the hand from the 4 possible cards of that value or = 6. Then for the card that appears 2 as the single you pick from the 4 possible cards of that value.thus the probability of getting such a hand is: ( = (c three of a kind? (This occurs when the cards have denominations a, a, a, b, c, where a, b, c are all distinct. ( 1 Again we choose the three values to appear in the hand. Then we choose from those the one to be the tripple ( choices. Now we have decided which is to be the triple and we choose the of the possible 4 cards of that denomination = 4. And for each of the values that appears as a single we choose from the four cards of that value or = 16. Consequently the probability is: 12 ( Problem. There are 0 physicists and 2 chemists attending a certain conference. Three of these people are randomly chosen to take part in a panel discussion. What is the probability that at least one chemist is chosen? 2
3 You could do this by cases considering the case when one chemist is chosen, where 2 chemists are chosen or where chemists are chosen. However I would rather consider the compliment of this event where no chemist is chosen and then subtract from one, the probability of this event happening. The total number of possible choices for a group of chosen from the total people is ( 0 where no chemist is chosen is (. The total number of groups. Thus the probability of no chemist is: = Thus the probability of getting at least one chemist is 1 minus the number above or: Problem 4. There are hotels in a certain town. If people check into hotels in a day, what is the probability that they each check into a different hotel? We could view this problem as the intersection of the following three events. E 1 = {event person 1 checks into a hotel}, E 2 = {event that person 2 doesn t check into the same hotel as person 1}, E = {event that person is not in hotels with either person one or person 2}. We would be interested in P (E 1 E 2 E = P (E 1 P (E 2 E 1 P (E E 1 E 2. The E 1 is the entire sample space since person 1 must check into SOME hotel, so P (E 1 = 1. Once person one has checked into a hotel if person 2 cant check into the same one then he has 4 choices out of where he or she can check in thus P (E 2 E 1 = 4. Once person one and person 2 are checked into different hotels then there is only choices left for where person can check in if he is not to be in the same hotel as person 1 or person 2. Thus P (E E 1 E 2 =. Consequently, the probability that we are after is equal to 4 = Problem. (a If n people, including A and B, are randomly arranged in a line, what is the probability that A and B are next to each other? Well the sample space has a total of n! possible orderings. Then we count the ways in which A and B can be next to each other. To do this we consider them as one person and order the n1 people in (n 1! ways. Then we must order A and B among themselves in 2 possible ways. Thus the probability of selecting such an ordering would be 2 (n 1! n! = 2 n. (b What would the probability be if the people were randomly arranged in a circle?
4 If they are arranged in a circle, then as argued before, the number of ways to do that is (n 1!. The number of ways in which A and B are next to each other can be counted by fixing A at the 12 o clock position on the table and then placing B either to his left or right (2 choices and then ordering the remaining (n 2! possible ways. Thus this probability would be 2(n 2! = 2. (n 1! n 1 Problem 6. Find the simplest expression for the following events: (a (E F (E F c = E; (b (E F (E c F (E F c = F E; (c (E F (F G = F EG. Problem 7. In this problem we prove the following identity in two different ways. (2k! 2 k k! = (2k 1(2k (2k...1. (1 (a Prove the above identity using induction. To prove by induction we must do two things. First we must prove the base case when k=1. To do this we plug in 1 into both sides of the equation. Both sides give 1, thus the equation holds when k=1. Now we must assume that the formula holds for k = n and prove that this implies that the formula holds for k = n + 1. To do this we write down the left hand side of the equation replacing k with n + 1 giving: (2n + 2! 2 n+1 (n + 1! = (2n + 2(2n + 1(n! 2(n + 12 n n! = (2n + 1 [ (2n! 2 n n! Now the fraction in the square brakets in the equation above is exactly the LHS of the equation when k = n. We are assuming that the equation holds when k = n so we can replace this term in the square brackets with the right hand side of the equation when k = n. Thus we have that: ] (2 (2n + 2! = (2n+1(2n 1(2n (2kn...1 = (2(n+1 1(2(n n+1 (n + 1! ( Thus the equation holds for k = n + 1 if it is true for k = n. Consequently since it is true for the base case of k = 1 then it is true for all integers k by induction as desired. (b Prove the above identity combinatorially. To do this consider the way to divide 2k people up into pairs. 4
5 To prove combinatorially we count the number of ways to split 2k people into k pairs. One way of counting this is to first consider if the k pairs were distinct in some way, (this is like (a from homework #1. The number of ways of doing this ( 2k is Where there are k 2 s. But then we must divide by k! since we don t 2, 2, 2,...2 want to have the pairs ordered. Doing this gives the formula on the left hand side of the equation. Another way of counting the same thing is to envision the 2k people in a room. The first person then chooses a partner. They have (2k 1 for their partner. Now those two people are gone and we go to the next remaining person, and he chooses a partner, and he chooses a partner from the remaining 2k people. this continues. Thus the number of ways of forming groups is (2k 1(2k...1 or the right hand side of the equation. Since both sides of the equation count the same thing then they must be equal. Problem 8. Let A B, and P (A = 0., P (B = 0.4. Calculate the following probabilities: P (A B, P (A B c, P (B A, P (B A c, P (B c A c. Show your calculations in detail. Solution First note that on one hand P (A B = P (A + P (B P (AB. On the other hand since A B then P (A B = P (B. Putting these two statements together we have that P (AB = P (A =.. P (A B = P (AB P (B =..4 =.7 P (A B c = P (ABc P (B c = 0 1 P (B = 0 (since A B, then ABc = empty set P (B A = P (AB P (A =.. = 1 P (B A c = BAc P (A c = P (B P (AB 1 P (A =.1.7 = 1 7. (note that P (B = P (BAc + P (BA P (B c A c = P (Bc A c = P (Bc = 1.4 =.6 = 6. (Note that since A B, then P (A c P (A c B c A c and thus A c B c = B c. Problem 9. Suppose that an ordinary deck of 2 cards (which contains 4 aces is randomly divided into 4 hands of 1 cards each. We are interested in determining p, the probability that each hand has an ace. Let E i be the event that the ith hand has exactly one ace, and E = E 1 E 2 E E 4 be the event that each hand has exactly one ace. (a Find P (E without using the multiplication rule simply count the number of ways 2 cards can be divided in 4 groups of 1 cards in each group, if each group contains exactly one ace. Please specify clearly whether you think of the hands as distinct
6 ( East, West, North, South hand or not (this will change the intermediate results, but not the final answer. ( 2 If we consider the players as distinct then there are a total of ways to 1, 1, 1, 1 hand out the cards. Count the number of such hands such that each player has eexactly one ace would be counted by 4! ways to hand out the aces, then ( 48 12, 12, 12, 12 ways to hand out the remaining non aces. Thus a probability of: 4! 48! 12!12!12!12! 2! 1!1!1!1! = (b Determine P (E by using the multiplication rule. If I think of the event E i that person i has exactly one ace Then I need P (E 1 E 2 E E 4. By the multiplication rule we have: P (E 1 E 2 E E 4 = P (E 4 E 1 E 2 E P (E E 2 E 1 P (E 2 E 1 P (E 1 8 ( 2 If we calculate P (E 1 then there are 4 out of all possible hands that 12 1 have exactly one ace. Similarly, if E 1 has once ace, then if person 2 has once ace, ( 6 then there are remaining choices for the ace and for the other 12 cards out of 12 ( 9 the possible hands that they could have if the 1 cards that person 1 has are 1 out of the deck. Similarly we get that: ( 24 ( 6 8 P (E 1 E 2 E E 4 = ( ( 26 ( 9 ( (c Show that your answers to (a and (b are the same. Simple computations reveal that these are the same. Problem 10. A deck of cards is shuffled and then divided into two halves of 26 cards each. A card is drawn from one of the halves; it turns to be an ace. This ace is then placed in the second halfdeck. The half is then shuffled, and a card is drawn from it. Compute the probability that this drawn card is an ace. 6
7 Hint: Condition on whether or not the interchanged card is selected. As the hint suggests we condition on whether or not the interchanged card is selected. Define the events A = { An Ace is selected on the second draw} and let I = {The interchanged card is selected}. Now we can write: P (A = P (IP (A I + P (I C P (A I C Since there are 27 cards in the pile after the interchanged card is placed in them, P (I = 1/27 and P (I C = 26/27. Now if we select the interchanged card the probability that I get an Ace is 1 since the interchanged card is an ace, thus P (A I = 1. Finally, if I select another card other then the interchanged card it should be equally likely to be any of the other 1 cards in the deck, of which are aces so we have that P (A I C = /1. Thus we have, P (A = (1/27(1 + (26/27(/1 = Problem 11. Suppose that an insurance company classifies people into one of the following three classes: low risk, average risk, and high risk. Their records indicate that the probabilities that a low, average, and high risk persons will be involved in an accident over a 1year period are 0.0, 0.1, and 0.0, respectively. Let 0% of the population be lowrisk, 0% be averagerisk, and 20% be highrisk. (a What proportion on people have accidents in a fixed year? Let A be the event that someone has an accident in a fixed year. Let L, V, and H be the events thats a certain person is Low, Average, or High risk respectively. In this question we are seeking to find P (A P (A = P (HP (A H+P (LP (A L+P (V P (A V = (.2(.+(.(.0+(.(.1 =.1 (b If Ms. Barry had no accidents in 200, what is the probability that she is a lowrisk policyholder? Averagerisk? What we are seeking here is P (L A c and P (V A c P (L A c = P (LP (Ac L P (A c P (V A c = P (V P (Ac V P (A c = (.(.9 1 P (A = = (.(.8.8 Problem 12. A truefalse question is to be posed to a husband and wife team on a quiz 7 =.
8 show. Both the husband and the wife will, independently, give the correct answer with probability p. Which of the following is a better strategy for this couple? (Give your calculations in detail for both strategies. Choose one of them and let that person answer the question; or In this situation we break the problem down be defining the following events. C is the event that they submit the correct answer. W is the event that the wife is chosen to answer. H is the probability that the husband is chosen to answer. (a P (C = P (CW + P (CH = P (W P (C W + P (HP (C H = p p1 2 = p. Here we are assuming that the husband and wife are chosen to answer the question equally likely. (b Have them both consider the question and then either give the common answer if they agree or, if they disagree, flip a coin to determine which answer to give? In this case we break up the problem in the following disjoint events, CC is the event that both answer correctly, II is the event that they both answer incorrectly. IC is the event that the husband answers incorrectly and the wife correctly, and CI is the event that the husband answered correctly and the wife incorrectly. Thus: P (C = P (CCP (C CC + P (IIP (C II + P (ICP (C IC + P (CIP (C CI. Now P (C CC = 1, P (C II = 0 and P (C IC = P (C CI = 1 2. Then substituting we get: P (C = p 2 (1 + (1 p 2 (0 + (1 p(p (1 p(p 1 2 = p Thus either strategy will be equally as good. 8
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