Math/Stat 394 Homework 2

Transcription

1 Math/Stat 39 Homework Due Wednesday Jan Six awards are to be given out among 0 students. How many ways can the awards be given out if (a one student will win exactly five awards. (b one student will win exactly four awards. (c no student can win more than three awards. ( (a There are 0 choices for which student wins five awards, 6 choices for which awards they win and 39 choices for who wins the remaining award. Thus there are ( ways the awards can be given out such that a student can win exactly five awards. ( (b There are 0 choices for which student wins five awards, 6 choices for which awards they win and 39 choices for who wins the remaining award. Thus there are ( ways the awards can be given out such that a student can win exactly four awards. (c There are 0 6 total number of choices of how to give out the six awards. We subtract off the mumber of ways that one student wins six awards (0, that one student wins five awards 1

2 (0 ( 6 (39 and the number of ways one student can win four awards 0 ( 6 39 to get ( ( ( ways the awards can be given out such that a student can win at most three awards.. A bridge hand consists of 13 cards from a standard card deck. How many bridge hands are there with (a exactly spades? (b cards if one suit, three cards of two suits and two cards of the fourth suit? (c 10 cards of one suit? (a There are ( ( 13 choices of the spades and 39 8 choices of the 8 other cards. Thus there are ( 13 ( 39 8 total hands with exactly spades. (b There are twelve choices for which suits get which numbers (,1,1. There are ( ( ( 13 ( 3 3 and 13 choices for which ranks are chosen in the different suits. Thus there are ( ( ( ( total choices. (c There are choices of the suit which gets 10 cards, ( choices of the cards in that suit and ( 39 3 choices of the remaining 3 cards. Thus there are ( ( total choices. and from chapter of Ross problems 3,,6,9,1,17 and 37 (which are copied below.

3 3. Two dice are thrown. Let E be the event that the sum of the dice is odd; let F be the event that at least one of the dice lands on 1; and let G be the event that the sum is five. Describe the events EF, E F, F G, EF C and EF G EF is the event that one dice lands on 1 and the other lands on an even number so the sum is odd. E F is the event that either the sum of the two dice is odd or one die lands on 1 and the other die is odd. F G is the event that one die lands on 1 and the sum is five so the other lands on. EF C is the event that the sum of the dice is odd and neither lands on 1. EG is the same as G (if the sum is five then it is odd so EF G = F G which is the event that one die lands on 1 while the other lands on.. A system is composed of components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector (x 1, x, x 3, x, x, where x i is equal to 1 if component i is working and is equal to 0 if component i is failed. (a How many outcomes are in the sample space. (b Suppose that the system will work if components 1 and are both working, or if components 3 and are both working, or if components 1,3 and are all working. Let W be the event that the system will work. Specify all the outcomes in W. (c Let A be the event that components and are both failed. how many outcomes are contained in the event A? (d Write out all the outcomes in the event AW. (a 3. Each of five components can be either 0 or 1. (b W consists of all eight vectors where both components 1 and are working (1, 1, 0, 0, 0 (1, 1, 0, 0, 1 (1, 1, 0, 1, 0 (1, 1, 0, 1, 1 (1, 1, 1, 0, 0 (1, 1, 1, 0, 1 (1, 1, 1, 1, 0 and (1, 1, 1, 1, 1. It also consists of six additional elements where both components 3 and are working (0, 0, 1, 1, 0 (0, 0, 1, 1, 1 (0, 1, 1, 1, 0 (0, 1, 1, 1, 1 (1, 0, 1, 1, 0 (1, 0, 1, 1, 1 and one element where components 1,3 and are working. (1, 0, 1, 0, 1 (c 8. A is the event that both components and have failed so components 1, and 3 are free to be either working or failed. (d The outcomes in the event AW are (1, 1, 0, 0, 0 and (1, 1, 1, 0, 0. 3

4 6. A hospital administrator codes incoming patients suffering gunshot wounds according to whether they have insurance (coding 1 if they do and 0 if they do not and according to their condition, which is rated as good (g, fair (f or serious (s. Consider an experiment that consists of determining the type of the coding of such a patient. (a Give the sample space of the experiment. (b Let A be the event that the patient is in serious condition. Specify the outcomes in A. (c Let B be the event that the patient is uninsured. outcomes in B. (d Give all the outcomes in the event B C A. (a S = {(i, j : i {0, 1} and j {g, f, s}}. (b A = {(0, s, (1, s} (c B = {(0, g, (0, f, (0, s} (d B C A = {(1, s} Specify the 9. A retail establishment accepts either the American Express or the VISA credit card. A total of percent of its customers carry an American Express card, 61 percent carry a VISA card, and 11 percent carry both. What percent of its customers carry a credit card the establishment will accept? Let V be the event that the customer carries VISA and A be the event that they carry American Express. We want to find P (A V. From the problem we see that P (A =., P (V =.61 and P (A V =.11 and plug it into the equation P (A V = P (A + P (V P (A V to get P (A V = = An elementary school is offering 3 language classes: one in Spanish, one in French and one in German. These classes are open to any of the 100 students in the school. There are 8 students in the Spanish class, 6 in the French class, and 16 in the German class. There are 1 students in both Spanish and French, that are in both Spanish and German, and 6 that are in both French and German. In addition, there are students taking all 3 classes.

5 (a If a student is chosen randomly, what is the probability that he or she is not in any of these classes? (b If a student is chosen randomly, what is the probability that he or she is taking exactly one language class? (c If students are chosen randomly, what is the probability that at least one is taking a language class. Let S be the event that a student is taking Spanish, let G be the event that a student is taking German and let F be the event that a student is taking French. (a We can use inclusion exclusion to find the probability a student is taking at least one class. P (S G F = P (S + P (G + P (F P (SG P (SF P (F G + P (SF G = =.. Then P ((SF G C = 1 P (SF G =. so there are 0 taking at least one class. (b There are 10 students in Spanish and French but not German, students in Spanish and German but not French, and students in German and French but not Spanish. Thus there are 16 students taking exactly two classes and two taking all three. Thus there are 3 students taking exactly one language class and P (a student is taking exactly one class =.3. (c There are ( ( 100 choices of two students. Of those 0 have both students not taking a language class so ( ( have at least one student taking a language class. Thus the probability that at least one of the two is taking a language class is ( 0 ( 100 ( If eight castles (that is rooks are randomly placed on a chessboard compute the probability that none of the rooks can capture any of the others. That is, compute the probability that no row or file contains more than one rook.

6 If we have already placed down i rooks with no two rooks in the same row or column then there are 8 i possible choices for the row and 8 i possible choices for the column so there are (8 i possible choices for where to put the i + 1st rook. There are 6 possible squares to place the first rook, 9 places to place the second, 36 for the third, for the fourth all the way to 1 square for the eighth. Thus there are (6(9(36((16(9((1 ordered ways to arrange the rooks. (The problem isn t specific we will assume that no two rooks are allowed on the same space. There are (6(63(6(61(60(9(8(7 possible ordered ways to arrange the rooks. Thus the probability that no two rooks are in the same row or column is (6(9(36((16(9((1 (6(63(6(61(60(9(8(7 37. An instructor gives her class a set of 10 problems with the information that the final exam will consist of a random selection of of them. If a student has figured out how to do 7 of the problems, what is the probability that he or she will answer correctly (a all five problems; (b at least four of the problems? (a There are ( 10 possible exams the instructor can give. There are exams that consist only of the seven problems the student has ( 7 figured out. Thus P (the student has figured out all problems = ( 7 ( 10. (b First we calculate the number of possible exams where the student has figured out exactly four problems. There are ( 7 choices of problems the student has figured out. For each of these there 3 choices of the problem the student hasn t figured out. Adding in the ( 7 possible exams that the student figured out five problems we get that there are ( ( possible exams where the student has figured out at least four problems. Thus P (the student has figured out at least problems = 3( ( ( 10. 6