Homework 2 Sample Problems and Solution
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North Carolina State University *
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724
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Mathematics
Date
Jun 19, 2024
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14
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ISEN 370 Homework 2 Sample Questions and Solutions 1. List the elements of each of the following sample spaces. Complete parts (a) through (e) below. (a) The set of integers between 1 and 50 divisible by 6. The set of all possible outcomes of a statistical experiment is called the sample space and is represented by the symbol S. The sample space for this set is all integers that satisfy the given condition. The set of integers between 1 and 50 that are divisible by 6 is the same as the multiples of 6. The set that represents the multiples of 6 between 1 and 50 is S={6,12,18,24,30,36,42,48}. (b) The set 𝑆𝑆
= {
𝑥𝑥
|
𝑥𝑥
2
+ 2
𝑥𝑥 −
15 = 0}
. 𝑆𝑆
= {
𝑥𝑥
|
𝑥𝑥
2
+ 2
𝑥𝑥 −
15 = 0}
. Sample spaces with a large or infinite number of sample points are often described by a statement or rule method. The vertical bar in the rule is read “such that”. For this set, the rule is 𝑥𝑥
2
+ 2
𝑥𝑥 −
15 = 0
Solve the equation for x (x+5)(x-3)=0 x = -5,3 Thus, the set 𝑆𝑆
= {
𝑥𝑥
|
𝑥𝑥
2
+ 2
𝑥𝑥 −
15 = 0}
contains the elements -5 and 3. (c) The set of outcomes when a coin is tossed until a tail or two heads appear. Let H represent a head and T represent a tail. List the possible outcomes of the first two tosses. HT, HH, T List all of the outcomes for the set. S={HT,HH,T} (d) The set S= {x | x is a continent}
For this set, the rule is that x must be a continent. List the continents. S = {N. America, S. America, Europe, Asia, Africa, Australia, Antarctica} (e) The set 𝑆𝑆
= {
𝑥𝑥
|4
𝑥𝑥 −
20
≥
0 𝑎𝑎𝑎𝑎𝑎𝑎
𝑥𝑥
< 4}
. The rule for this set is 4
𝑥𝑥 −
20
≥
0 𝑎𝑎𝑎𝑎𝑎𝑎
𝑥𝑥
< 4
Solve the inequality 4x
−
20
≥
0 for x. 4x
−
20 ≥
0 x ≥
5 Therefore, the rule for this set is x
≥
5 and x<3. There are no real numbers that satisfy both of these inequalities. Therefore, the set S is the empty set. S=Φ
2. An experiment involves tossing a pair of dice, one green and one red, and recording the numbers that come up. If x equals the outcome on the green die and y equals the outcome on the red die, describe the sample space S. Complete parts (a) and (b) below. (a) Describe the sample space S by listing the elements (x,y). X: 1,2,3,4,5,6 Y: 1,2,3,4,5,6 The elements (x,y) will have 36 combinations as shown below: S {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} (b) Describe the sample space S by using the rule method X takes value between 1 and 6, y takes value between 1 and 6. S= { (x,y) | 1 ≤ x ≤ 6 and 1 ≤ y ≤ 6}
3. An experiment consists of tossing a die and then flipping a coin twice if the number on the die is 4 or greater. If the number on the die is 3 or less, the coin is flipped once. Using the notation
H, for example, to denote the outcome that the die comes up and then the coin comes up heads, and HT to denote the outcome that the die comes up followed by a head and then a tail on the coin, construct a tree diagram to show the 18 elements of the sample space S. Tree diagram is shown below: 4. Two jurors are selected from 4 alternates to serve at a murder trial. Using the notation A1A3, for example, to denote the simple event that alternates 1 and 3 are selected, list the 6 elements of the sample space S. Answer: The set of all possible outcomes of a statistical experiment is called the sample space and is represented by the symbol S. Each outcome in a sample space is called an element or a member of the sample space, or simply a sample point. If the sample space has a finite number of elements, list the members separated by commas and enclosed in braces. The outcome A1A2 is possible since it represents the event that alternates 1, and 2 are selected. The outcomes A1A2 and A2A1 shouldn't both be included in the sample space because the order in which alternates are selected is not significant. The outcome A1A1shouldn't be included in the sample space because an alternate cannot be repeatedly selected.
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Since there are a total of 6 outcomes in the sample space, list all the outcomes separated by commas and enclosed in braces. S={A1A2, A1A3, A1A4, A2A3, A2A4, A3A4} 5. An experiment consists of tossing a die and then flipping a coin twice if the number on the die is 4 or greater. If the number on the die is 3 or less, the coin is flipped once. the notation 4H, for example, to denote the outcome that the die comes up 4 and then the coin comes up heads, and 3HT to denote the outcome that the die comes up 3 followed by a head and then a tail on the coin. Complete parts (a) through (e) below. Develop a tree diagram (a) List the elements corresponding to the event A that a number greater than 4 occurs on the die. A={5HH,5HT,5TH,5TT, 6HH, 6HT,6TH,6TT} (b) List the elements corresponding to the event B that a head then a tail occurs B={4HT,5HT,6HT} (c) List the elements corresponding to the set A′
A’={1H,1T,2H,2T,3H,3T,4HH,4HT,4TH,4TT} (d) List the elements corresponding to the set B. A′ ∩B
A′ ∩B
={4HT}
(e) List the elements corresponding to the set A
Ս
B. A
Ս
B={4HT, 5HH,5HT,5TH,5TT, 6HH, 6HT,6TH,6TT } 6. An engineering firm is hired to determine if certain waterways are safe for fishing. Samples are taken from two rivers. Complete parts (a) through (c) below. (a)
List the elements of a sample space S, using the letters F for safe to fish and N for not safe to fish. S={FF, FN,NF, NN} (b) List the elements of S corresponding to event E that at least one of the rivers are safe for fishing. E={FF,FN,NF} (c) Define an event that has as its elements the points {FF, NF} The second river was safe for fishing. 7. Construct a Venn diagram to illustrate the possible intersections and unions for the following events relative to the sample space consisting of all automobiles made in the United States. B: Blind spot warning P: Power steering R: Reverse camera 8. If S={0,1,2,3,4,5,6,7,8,9} and A={1,3,5,7,9}, B={0,2,4,6,8}, C={2,3,4,5} and D={0,7,8}, list the elements of the sets corresponding to the following events: (a) List the elements of the set corresponding to A
ꓴ
C. (b) List the elements of the set corresponding to A
ꓵ
B. (c) List the elements of the set corresponding to C’.
(d) List the elements of the set corresponding to (C’
ꓵ
D) ꓴ
B (e) list the elements of the set corresponding to (S ꓵ
C)’ (f) List the elements of the set corresponding to A
ꓵ
C
ꓵ
D 9. Let A, B, and C be events relative to the sample space S. Using the Venn Diagrams, select the regions corresponding to the following events (a) A ꓵ
C A ꓵ
C=regions 3,8 (b) (B
ꓴ
C)’ (B
ꓴ
C)’= Regions 1,2 (c) (A
ꓵ
B) ꓴ
C (A
ꓵ
B) ꓴ
C=3,4,5,7,8 10. Suppose that a family is leaving on a summer vacation in their camper and M is the event that they will experience mechanical problems, T is the event that they will receive a ticket for committing a traffic violation, and V is the event that they will arrive at a campsite with no vacancies. Referring to the accompanying Venn diagram, list the numbers of the regions that represent the following events. Complete parts (a) through (d) below.
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(a) The family will experience mechanical problems and will not receive a ticket for a traffic violation but will arrive at a campsite with no vacancies. 2 (b) The family will experience neither mechanical problems nor trouble in locating a campsite with a vacancy but will receive a ticket for a traffic violation. 7 (c) The family will either have mechanical trouble or arrive at a campsite with no vacancies and will receive a ticket for a traffic violation. 1,3,4 (d) The family will not experience mechanical problems. 3,6,7,8 11. Students at a private liberal arts college are classified as being freshmen, sophomores, juniors, or seniors, and also according to whether they are male or female. Find the total number of possible classifications for the students of that college. There are __8__total possible classifications for the students of the college. The first operation (classifying students based on class level) can be done in 4 ways, and the second operation (classifying students based on gender) can be done in 2 ways. Therefore, there are 4*2=8 total ways. 12. A study concluded that following 7 simple health rules can extend a man's life by 11 years on the average and a woman's life by 7 years. These 7 rules are as follows: no smoking, get regular exercise, use alcohol only in moderation, get 7 to 8 hours of sleep, maintain proper weight, eat breakfast, and do not eat between meals. In how many ways can a person adopt 5 of these rules to follow: (a) if the person presently violates 4 of the 7 rules?
(b) if the person never smokes and always exercises daily? (a) The person presently violates 4 out of the 7 rules, meaning the person adopts 3 rules presently and must adopt 2 new rules out of 4 rules. Since it does not matter what order the person adopts the rules in, the combination formula is used, where r is the total number of rules being newly adopted and n is the total number of presently violated rules. �
𝑎𝑎
𝑟𝑟
�
=
𝑎𝑎
!
𝑟𝑟
! (
𝑎𝑎 − 𝑟𝑟
)!
Substitute the values of n and r into the combination formula. n=4, r=2 �
𝑎𝑎
𝑟𝑟
�
=
𝑎𝑎
!
𝑟𝑟
! (
𝑎𝑎 − 𝑟𝑟
)!
=
4!
2! 2!
= 6
(b) In this case, we know that 2 of the rules are already being followed. Therefore, 3 of the remaining 5 rules must be followed in order to follow a total of 5. Set up the combination formula using n=5 and r=3. �
𝑎𝑎
𝑟𝑟
�
=
𝑎𝑎
!
𝑟𝑟
! (
𝑎𝑎 − 𝑟𝑟
)!
=
5!
3! 2!
= 10
13. In how many different ways can a true-false test consisting of 5 questions be answered? If an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two a third operation can be performed in n3 ways, and so forth, then the sequence of k operations can be performed in n1n2...nk ways. Each question is a single operation. Taking the test is a sequence of 5 questions. For each true-false question, there are 2 possible answers, so there are 2 ways each operation can occur. So the test is a sequence of 5 operations that can each occur 2 different ways. To find the number of ways the test can be answered, multiply 5 factors of 2. 2•2•2•2•2=32 14. A witness to a hit-and-run accident told the police that the license number contained the letters MHL followed by 3 digits, the first of which was a 5. If the witness cannot recall the last 2
digits, but is certain that all 3 digits are different, find the maximum number of automobile registrations that the police may have to check. If an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two a third operation can be performed in n3 ways, and so forth, then the sequence of k operations can be performed in n1n2...nk ways. In this case, because the witness can remember that the license number contains MHL5 followed by two digits and is certain that all 3 digits are different, there are 9 options for the next digit and 8 options for the last digit. Multiply the values together to find the maximum number of automobile registrations that the police may have to check. 9*8=72 15. (a) In how many ways can 5 people be lined up to get on a bus? (b) If 3 specific persons, among 5 insist on following each other, how many ways are possible? (c) If 2 specific persons, among 5, refuse to follow each other, how many ways are possible? (a) A permutation is an arrangement of all or part of a set of objects. The number of permutations of n objects is n!. Find n, the number of objects being arranged. n=5 Next, find the number of permutations of the n objects. 5!= 5*4*3*2*1=120 Thus, the 5 different people can be lined up in 120 different ways. (b) Recall that if an operation can be performed in n1 ways, and if for each of these ways a second operation can be performed in n2 ways, then the two operations can be performed together in n1n2 ways. So in this case, make n1 equal to the number of ways in which the 3 people who want to be together can be arranged, and n2 equal to the number of ways the other 2 people who have no preference can be arranged..n1=3, 3!=6 Next find n2, the number of ways the other 2 people who have no preference can be arranged. n2=2!=2 Now consider all 5 people lined up to get on the bus in places labeled 1 through 5, with place 1 being the first person on the bus, place 2 being the second person on the bus, and so on. Determine the number of different places the first person in the group of 3 people who want to be together can get on the bus.
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Note that logically, the first person in the group cannot get on the bus in such a place where there are fewer places than people remaining in the group. Therefore, there are 3 different places the first person in the group of 3 people can get on the bus. Lastly, multiply these values to find the total number of ways that they can line up. n1n2=6*2*3=36 Alternately, you can consider the three person as a unit, this unit and the rest of 2 people can be arranged in 3! Ways, and the three person unit itself can be arranged in 3! Hence, the total ways are 3!*3!=36 Therefore, the people can be lined up in 36 different ways. (c) To find the number of arrangements where these 2 people are not lined up next to each other, subtract the number of arrangements where 2 specific people are next to each other from the total number of arrangements. If we define the event A as these 2 people are lined up next to each other, then these 2 people are not lined up next to each other is A’. Therefore the number of arrangement these 2 people are not lined up next to each other is equal to total number of arrangement the 5 people can be arranged without any restriction minus the number of arrangements these 2 people are lined up next to each other. To determine the number of arrangements where 2 specific people are next to each other, consider that these 2 people can be arranged in either order, and at any point in the line. The following expression is correct for finding this number of arrangements. 4!*2!=48 Subtract this from the total number of possible permutations, 5!=120 to find the total number of permutations where these 2 people are not next to each other in line. 120-48=72 16. If a multiple choice test consists of 4 questions, each of which with 4 possible answers of which only 1 is correct, (a) in how many different ways can a student check off one answer to each question? (b) in how many ways can a student check off one answer to each question and get all the answers wrong? (a) If an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two a third operation can be performed in n3 ways, and so forth, then the sequence of k operations can be performed in n1n2...nk ways.
In this case, a total of 4 operations are occurring in sequence when the test is taken. Each question can be answered in 4 different ways. So the test is a sequence of 4 operations that can each occur 4 different ways. To find the number of ways the test can be answered, multiply 4 factors of 4. 4*4*4*4=256 (b) Determining the number of ways to get all of the questions wrong is found similarly to the process for part (a). The one difference in this calculation is the number of ways that each question can be answered. So, we know that the number of questions on the test is not changing. Therefore, the number of operations occurring in sequence is still 4. In order to get every question wrong, each question must be answered in one of 3 ways. This means that to get every question on the test wrong is a sequence of 4 operations that can each occur 3 different ways. To find the number of ways the test can be answered, multiply 4 factors of 3 3*3*3*3=81 17. (a) How many three-digit numbers can be formed from the digits 0,1,2,3,4,5,and 6 if each digit can be used only once? (b) How many of these are odd numbers? (c) How many are greater than 330? (a) Recall that if an operation can be performed in n1 ways, and if for each of these a second operation can be performed in n2 ways, and for each of the first two a third operation can be performed in n3 ways, and so forth, then the sequence of k operations can be performed in n1n2...nk ways. Determine the number of ways in which the hundreds digit can be chosen. Note that 0 cannot be the leftmost digit in a number: 6 Next, determine the number of ways in which the tens digit can be chosen. Keep in mind that 0 can be used for this digit, however whichever number was used in the hundreds place cannot. 6 Similarly, determine the number of ways in which the ones digit can be chosen, keeping in mind that two digits have already been used. 5 Multiply the number of possibilities for each digit together to find the total number of possibilities. 6*6*5=180 (b) How many of these are odd numbers?
In order for the number to be odd, the ones digit must be an odd number. There are 3 odd numbers to choose from. Thus there are 3 options for the ones digit. Once again, 0 cannot be used as the hundreds digit. Determine how many digits can be used in the hundreds place. 5 There are 5 digits left that can be used as the tens digit. Multiply the number of possibilities for each digit together to find the total number of three-digit numbers that are odd. 5*5*3=75 (c) How many are greater than 330? To find this number, consider two scenarios. Either the hundreds digit must be greater than 3, or the hundreds digit must be equal to 3 and the tens digit must be greater than 3. Notice, it is not allowed to have the hundredth and the tenth to be the same number (e.g., 33x). Begin with the first scenario. There are 3 options 4,5, or 6) for the hundreds digit that are greater than 3. With 1 digit being used for the hundreds place, 6 digits can be used in the tens place. Lastly for this scenario,5 digits can be used in the ones place. Multiply the number of possibilities for each digit together to find the total number of possibilities for this scenario. 3*6*5=90 Now consider the other scenario. In this scenario, it is known that the hundreds place must be a 3, so there is only 1 option for the hundreds digit. There are 3 options for the tens digit are there that are greater than 3. Lastly, there are 5 options remaining for the ones digit for this scenario after the hundreds and tens places are filled. Multiply the number of possibilities for each digit together to find the total number of possibilities for this scenario. 1•3*5=15 Add the possibilities for each of the two scenarios together to determine the total number of three-digit numbers can be formed that are greater than 330. 90+15=105 18. In how many ways can 5 starting positions on a basketball team be filled with 7 players who can play any of the positions? Permutation use applies to the scenario presented in the problem statement, because each starting position is unique and order matters.
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A permutation is an arrangement of all or part of a set of objects. The number of permutations of n objects is n!. The number of permutations of n distinct objects taken r at a time is n!/(n
−
r)!. The number of permutations of n objects arranged in a circle is (n
−
1)! The number of distinct permutations of n things of which n1 are of one kind, n2 of a second kind, ..., nk of a kth kind is n!/(n1!n2!...nk!). The scenario presented in the problem statement involves finding the number of permutations of n distinct objects taken r at a time. Identify n, the number of distinct objects, in this scenario. n=7 Identify r, the size of the group of objects taken. r=5 The number of permutations of n distinct objects taken r at a time is nPr=n!(n
−
r)!. Evaluate n!(n
−
r)! for n=7 and r=5. 7!(7
−
5)!=2520 Thus, the 5 starting positions on a basketball team can be filled with 7 players who can play any of the positions in 2520 ways. 19. In how many different ways can 7 cacti be planted in a circle? The scenario presented in the problem statement involves finding the number of permutations of n objects arranged in a circle. Identify n, the number of objects, in this scenario. n=13 The number of permutations of n objects arranged in a circle is (n
−
1)!. Evaluate (n
−
1)! for n=7. (7
−
1)!=6!=720 Thus, there are 720 different ways in which 7cacti can be planted in a circle. 20. In how many ways can 5 oaks, 6 pines, and 4 maples be arranged along a property line if one does not distinguish among trees of the same kind? Permutation use applies to the scenario presented in the problem statement. A permutation is an arrangement of all or part of a set of objects. The number of permutations of n objects is n!. The number of permutations of n distinct objects taken r at a time is mnPr=n!/(n
−
r)!. The number of permutations of n objects arranged in a circle is (n
−
1)!
The number of distinct permutations of n things of which n1 are of one kind, n2 of a second kind, ..., nk of a kth kind is n!/(n1!n2!...nk!). The scenario presented in the problem statement involves finding the number of distinct permutations of n things of which n1 are of one kind, n2 of a second kind, ..., nk of a kth kind. Let n1 be the number of oaks. Identify n1.n1=5 Let n2 be the number of pines. Identify n2. n2=6 Let n3 be the number of maples. Identify n3. n3=4 Now determine n, the total number of trees, by adding the number of oaks, the number of pines, and the number of maples. 5+6+4=15 The number of distinct permutations of n things of which n1 are of one kind, n2 of a second kind, ..., nk of a kth kind is n!/(n1!n2!...nk!). (15!)/(5!*6!*4!)=630630 Thus, the trees can be arranged along a property line in 630630 ways if one does not distinguish among trees of the same kind.