Free math worksheets

- Which person is fixed in the first position? - How many people remain for the other two positions? - List the two possible orders of those people.

1. Lisa is fixed in the first position. 2. Tim and Max can fill the remaining two positions in \(2!=2\) orders. 3. The orders are Lisa–Tim–Max and Lisa–Max–Tim.

\(2\) orders: Lisa–Tim–Max and Lisa–Max–Tim

- Place the cards into the four digit positions one at a time. - How many cards are available before the first choice, and how does that number change? - Why does the fundamental counting principle multiply the successive numbers of choices?

a) There are \(4\) choices for the thousands digit, then \(3\) for the hundreds digit, \(2\) for the tens digit, and \(1\) for the ones digit. Thus, the total is \(4\cdot3\cdot2\cdot1=4!=24\). b) The fundamental counting principle multiplies the number of choices at successive stages, so the product \(4\cdot3\cdot2\cdot1\) counts every complete arrangement exactly once.

a) \(24\) numbers b) The successive numbers of choices are \(4,3,2,\) and \(1\), so \(4\cdot3\cdot2\cdot1=24\).

- How many permutations are there of \(n\) distinct objects? - Multiply the number of arrangements by the time for each arrangement. - Use \(3600\,\text{s}=1\,\text{h}\) to convert the final time.

a) The number of permutations of \(7\) distinct books is \(7!=7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1=5040\). b) The total time is \(5040\cdot10\,\text{s}=50{,}400\,\text{s}\). Converting to hours gives \(50{,}400\div3600=14\).

a) \(5040\) arrangements b) \(14\) hours

- Which cards can be used in the hundreds place if the number must be less than \(600\)? - After one card is used, how many remain for the next position? - Multiply the numbers of choices for the three positions.

1. For the number to be less than \(600\), the hundreds digit must be \(2\) or \(4\), giving \(2\) choices. 2. After choosing the hundreds digit, \(3\) cards remain for the tens digit. 3. Then \(2\) cards remain for the ones digit. 4. By the fundamental counting principle, the total is \(2\cdot3\cdot2=12\).

\(12\) numbers

- Count the choices for each position. - For part b, track how the available choices decrease after each digit is used. - For part c, count the codes in which one digit is repeated four times.

a) Each of the four positions has \(6\) choices, so there are \(6^4=1296\) codes. b) For four different digits, the successive choices are \(6,5,4,3\), giving \(6\cdot5\cdot4\cdot3=360\) codes. c) There are \(6\) codes in which one digit is repeated four times. Therefore, \(P(\text{all digits equal})=\frac{6}{1296}=\frac{1}{216}\approx0.00463\).

a) \(1296\) b) \(360\) c) \(\frac{1}{216}\approx0.00463\)

- Use a factorial to arrange distinct people in a row. - For part b, count only the people whose positions are not fixed. - For part c, arrange the first three people and the remaining five people separately, then multiply.

a) Eight distinct people can be arranged in \(8!=40{,}320\) ways. b) A and B are fixed in the first two positions in that order, so the remaining six people can be arranged in \(6!=720\) ways. c) A, B, and C can be arranged in the first three positions in \(3!\) ways, and the remaining five people can be arranged in \(5!\) ways. Thus, \(3!\cdot5!=720\).

a) \(40{,}320\) b) \(720\) c) \(720\)

- Count all arrangements of \(8\) distinct objects. - Count the arrangements that are sorted by page count. - Identify the assumption made about the different arrangements.

1. The \(8\) distinct books can be arranged in \(8!=40{,}320\) equally likely orders. 2. Exactly \(2\) orders satisfy the condition: increasing page count and decreasing page count. 3. Therefore, \(P=\frac{2}{8!}=\frac{1}{20160}\approx 0.00004960\). 4. This is a theoretical probability calculation based on equally likely outcomes.

The probability is \(\frac{1}{20160}\approx 0.00004960\). It is a theoretical probability based on equally likely arrangements.

- Decide whether the selected students have distinct roles. - Compare an ordered assignment with an unordered group. - For part b, count one specified group among all equally likely groups.

a) In case (1), the roles are distinct, so there are \(15\cdot14\cdot13=2730\) assignments. In case (2), only group membership matters, so there are \(\binom{15}{3}=455\) groups. b) Exactly one of the \(455\) equally likely groups is the specified group. Therefore, \(P=\frac{1}{455}\).

a) Case (1): \(2730\); case (2): \(455\) b) \(\frac{1}{455}\approx0.22\%\)

- Imagine all tickets arranged in one random order. - Ask whether any position is favored over another.

a) The grand-prize ticket is equally likely to be any of the \(100\) tickets drawn first, so the probability is \(\frac{1}{100}\). b) In a random ordering of all tickets, the grand-prize ticket is equally likely to occupy any position. Therefore, the probability that it is tenth is also \(\frac{1}{100}\). c) Every permutation of the tickets is equally likely, and the grand-prize ticket appears equally often in each position across all permutations.

a) \(\frac{1}{100}=0.01\) b) \(\frac{1}{100}=0.01\) c) Random ordering makes every position equally likely for the grand-prize ticket.

- Count all complete arrangements first. - Fix the required first position. - Remove the arrangements violating the second condition.

1. There are \(4!=24\) equally likely arrangements. 2. With A first, the remaining letters have \(3!=6\) arrangements. 3. Of those, B is last in \(2!=2\) arrangements, so \(6-2=4\) are favorable. The probability is \(\frac{4}{24}=\frac{1}{6}\).

The probability is \(\frac{1}{6}\).

- Count only numbers satisfying the distinct-digit condition. - The first digit cannot be zero. - Compare the constrained first-digit count with the full conditional sample space.

1. There are \(9\cdot9\cdot8=648\) valid three-digit numbers with distinct digits. 2. If the first digit is \(5\), the second digit has \(9\) choices and the third has \(8\), giving \(72\) favorable numbers. 3. Therefore, the conditional probability is \(\frac{72}{648}=\frac{1}{9}\).

The probability is \(\frac{1}{9}\).

- Compare the number of favorable arrangements with the total number of arrangements of the six distinct letters. - For part a, decide which positions are already determined before arranging the remaining letters. - For part b, think about how adjacency changes the number of independent positions while still allowing the two vowels to switch order.

a) There are \(6!=720\) total arrangements. Fix F in the first position and A in the second position; the remaining \(4\) letters can be arranged in \(4!=24\) ways. Therefore, \(P=\frac{24}{720}=\frac{1}{30}\). b) Regard A and O as one adjacent unit. That unit and the four consonants form \(5\) objects, giving \(5!=120\) arrangements. The vowels can occur as AO or OA, so there are \(2\cdot120=240\) favorable arrangements. Thus, \(P=\frac{240}{720}=\frac{1}{3}\).

a) \(\frac{1}{30}\) b) \(\frac{1}{3}\)

- First focus on the requirement that Ava and Ben occupy consecutive seats. - How could you count placements while keeping those two seats together as you arrange the other friends? - After fixing the adjacent pair as a unit, remember that the two people can exchange positions within it.

1. Regard Ava and Ben as one adjacent unit. Together with Clara, David, and Eli, there are \(4\) units to arrange. 2. The \(4\) units can be arranged in \(4!=24\) ways. 3. Ava and Ben can switch order within their adjacent unit, giving \(2\) internal orders. 4. Therefore, the number of seating arrangements is \(24\cdot2=48\).

\(48\) seating arrangements

- In part a, how many positions remain after the two specified books are fixed? - In part b, separate the two possible end positions for the math book. - For each possible end position, count the permutations of the other three books.

a) The math and history books are fixed at the two ends. The two remaining books can be arranged in \(2!=2\) ways. b) The math book can be at either end, giving \(2\) disjoint cases. In each case, the other three books can be arranged in \(3!=6\) ways. Therefore, the total is \(2\cdot3!=12\).

a) \(2\) arrangements b) \(12\) arrangements

- Count the total number of available characters. - Compare how the number of choices changes when repetition is allowed versus forbidden. - For part c, compare the number of all-digit codes with the total number of codes.

a) There are \(18\) available characters, and each of the five positions has \(18\) choices. Thus, \(18^5=1{,}889{,}568\). b) Without repetition, the successive numbers of choices are \(18,17,16,15,14\), so the number of codes is \(18\cdot17\cdot16\cdot15\cdot14=1{,}028{,}160\). c) There are \(10^5\) all-digit codes out of \(18^5\) total codes. Therefore, \(P(\text{all digits})=\frac{10^5}{18^5}=\left(\frac{5}{9}\right)^5=\frac{3125}{59049}\approx0.0529\).

a) \(1{,}889{,}568\) b) \(1{,}028{,}160\) c) \(\frac{3125}{59049}\approx 5.29\%\)

- Decide whether repetition is allowed in each part. - Decide whether order matters. - Use a product for ordered selections and a combination for an unordered group.

a) With repetition allowed and order recorded, each spin has \(20\) possible results, so there are \(20^3=8000\) sequences. b) With no repeated number, the successive choices are \(20,19,18\). Thus, there are \(20\cdot19\cdot18=6840\) sequences. c) When three distinct numbers are selected without regard to order, the number of groups is \(\binom{20}{3}=1140\).

a) \(8000\) b) \(6840\) c) \(1140\)

- Determine whether order matters in each part. - Determine whether a digit remains available after it is used. - Match each case to an ordered count or a combination.

a) With repetition allowed and order important, there are \(9^5=59{,}049\) codes. b) Without repetition and with order important, there are \(9\cdot8\cdot7\cdot6\cdot5=15{,}120\) codes. c) Without repetition and with order irrelevant, there are \(\binom{9}{5}=126\) sets.

a) \(59{,}049\) b) \(15{,}120\) c) \(126\)

- Determine whether the selected people have distinct roles. - Determine whether order matters in each situation. - Determine whether a name may be repeated.

a) The three roles are distinct and no student can be repeated, so there are \(25\cdot24\cdot23=13{,}800\) assignments. b) The four representatives form an unordered group, so there are \(\binom{25}{4}=12{,}650\) groups. c) Each of the four positions has \(25\) choices and repetition is allowed, so there are \(25^4=390{,}625\) sequences.

a) \(13{,}800\) b) \(12{,}650\) c) \(390{,}625\)

- Ask whether an item can be selected more than once. - Ask whether changing the order creates a different outcome. - Identify the number of available choices and the number of selections.

a) \(n=10\), \(k=4\), repetition is allowed, and order matters. The number of codes is \(10^4=10{,}000\). b) \(n=10\), \(k=3\), repetition is not allowed, and order matters. The number of outcomes is \(10\cdot9\cdot8=720\). c) \(n=12\), \(k=2\), repetition is not allowed, and order does not matter. The number of groups is \(\binom{12}{2}=66\).

a) \(n=10\), \(k=4\), repetition allowed, order matters; \(10{,}000\) outcomes b) \(n=10\), \(k=3\), no repetition, order matters; \(720\) outcomes c) \(n=12\), \(k=2\), no repetition, order does not matter; \(66\) outcomes

- Compare the restrictions in parts a) and b): one forbids any reuse, while the other only restricts neighboring positions. - For part c, first decide what must be true of the two digits used by an alternating code. - After choosing the two digits, consider how the first position determines the rest of the pattern.

a) Without repetition, the four positions have \(6,5,4,3\) choices, so there are \(6\cdot5\cdot4\cdot3=360\) codes. b) The first position has \(6\) choices. Each later position has \(5\) choices because it cannot equal the immediately preceding digit. Thus, there are \(6\cdot5^3=750\) valid codes. c) There are \(\binom{6}{2}=15\) pairs of distinct digits. Each pair produces two alternating codes, depending on which digit comes first, so there are \(30\) favorable valid codes. Therefore, \(P=\frac{30}{750}=\frac{1}{25}=0.04\).

a) \(360\) b) \(750\) c) \(\frac{1}{25}=4\%\)

- For part a, count the choices for each square. - For part b, track how the number of available colors changes after each choice. - For part c, compare the desired event with the event counted in part b.

a) Each of the five positions has \(12\) choices, so there are \(12^5=248{,}832\) patterns. b) For five different colors, the successive choices are \(12,11,10,9,8\), giving \(12\cdot11\cdot10\cdot9\cdot8=95{,}040\) patterns. c) Use the complement that all five colors are different. Thus, \(P(\text{at least one repeat})=1-\frac{95040}{248832}=\frac{89}{144}\approx0.6181\).

a) \(248{,}832\) b) \(95{,}040\) c) \(\frac{89}{144}\approx 61.81\%\)

- Write the binomial coefficient using factorials. - When comparing the two expressions, look for factors that cancel. - What does the remaining factor say about the number of orders of the selected \(k\) objects?

a) \(A=\frac{9!}{5!}=9\cdot8\cdot7\cdot6=3024\), and \(B=\binom{9}{4}=126\). b) \(\frac{A}{B}=\frac{3024}{126}=24=4!\). c) In general, \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\). Therefore, \(\frac{A}{B}=\frac{n!}{(n-k)!}\div\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!}\cdot\frac{k!(n-k)!}{n!}=k!\).

a) \(A=3024\), \(B=126\) b) \(\frac{A}{B}=24\) c) Substituting \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\) and canceling leaves \(k!\).

- Count the number of factors in the product. - Expand \(n!\) and identify the factors contained in \((n-k)!\). - Substitute \(k=n\) and use \(0!=1\).

a) When \(n=10\) and \(k=3\), the product has three factors: \(10\cdot9\cdot8=720\). b) Expand \(n!\) as \(n!=n(n-1)\cdots(n-k+1)(n-k)!\). Dividing by \((n-k)!\) leaves \(\frac{n!}{(n-k)!}=n(n-1)\cdots(n-k+1)\). c) If \(k=n\), then \(\frac{n!}{(n-n)!}=\frac{n!}{0!}=n!\) because \(0!=1\).

a) \(720\) b) \(\frac{n!}{(n-k)!}=n(n-1)\cdots(n-k+1)\) c) \(n!\)

- Start by counting the equally likely ordered flavor choices for all six teenagers. - For part a, track how many flavors remain available if every new choice must differ from earlier choices. - For part b, decide how many common flavors could produce the event.

a) There are \(10^6\) equally likely ordered sets of choices. For all different flavors, the successive numbers of choices are \(10,9,8,7,6,5\). Thus, \(P=\frac{10\cdot9\cdot8\cdot7\cdot6\cdot5}{10^6}=0.1512\). b) There are \(10\) favorable ordered outcomes in which everyone chooses the same flavor, one for each flavor. Thus, \(P=\frac{10}{10^6}=0.00001\).

a) \(0.1512\) b) \(0.00001\)

- For part a, think about how the four specified songs can remain consecutive while the whole playlist is rearranged. - Account separately for the placement of that consecutive group and the order of the songs within it. - For part b, focus on the membership of the first five positions rather than their internal order.

a) Regard the \(4\) band songs as one consecutive unit. That unit and the other \(8\) songs form \(9\) objects, which can be arranged in \(9!\) ways. The \(4\) band songs can be arranged internally in \(4!\) ways. Thus, the number of play orders is \(9!\cdot4!=8{,}709{,}120\). b) The first \(5\) songs form an unordered sample without replacement. Choose \(2\) of the \(4\) band songs and \(3\) of the \(8\) other songs out of all \(5\)-song subsets. Therefore, \(P=\frac{\binom{4}{2}\binom{8}{3}}{\binom{12}{5}}=\frac{14}{33}\approx0.4242\).

a) \(8{,}709{,}120\) play orders b) \(\frac{14}{33}\approx0.4242\), or about \(42.42\%\)

- List the pairs of positions that have exactly one position between them. - Account for the two possible orders of Anna and Ben. - For part b, count the left-end and right-end cases separately.

a) There are \(8\) position pairs with exactly one position between them: \((1,3)\) through \((8,10)\). Anna and Ben can switch places in \(2\) ways, and the other \(8\) people can be arranged in \(8!\) ways. Thus, the count is \(8\cdot2\cdot8!=645{,}120\). b) For both people in the first three positions, choose their two positions in \(\binom{3}{2}\) ways, arrange Anna and Ben in \(2!\) ways, and arrange the other \(8\) people in \(8!\) ways. The same count applies to the last three positions, and the two events are disjoint. Therefore, \(P=\frac{2\binom{3}{2}2!8!}{10!}=\frac{2}{15}\approx0.1333\).

a) \(645{,}120\) lineups b) \(\frac{2}{15}\approx0.1333\), or about \(13.33\%\)

- Determine whether the prizes make the order of winners important. - Use an unordered count for identical prizes and an ordered count for ranked prizes. - For one exact equally likely outcome, compare one favorable result with the full sample space.

a) In model (1), order does not matter because the prizes are identical, so there are \(\binom{20}{4}=4845\) winner groups. In model (2), the four places are distinct, so there are \(20\cdot19\cdot18\cdot17=116{,}280\) ordered outcomes. b) Each outcome within a model is equally likely. Therefore, the probabilities of one exact prediction are \(\frac{1}{4845}\approx0.000206\) for model (1) and \(\frac{1}{116280}\approx0.00000860\) for model (2).

a) Model (1): \(4845\); model (2): \(116{,}280\) b) Model (1): \(\frac{1}{4845}\approx0.0206\%\); model (2): \(\frac{1}{116280}\approx0.000860\%\)

- Count all length-\(n\) codes. - Count the codes that use every value exactly once. - Form a probability in terms of \(n\) and test small positive integers.

1. There are \(n^n\) possible codes. 2. A code with all different entries is a permutation of the \(n\) available values, so there are \(n!\) favorable codes. 3. Thus, \(P(\text{all different})=\frac{n!}{n^n}\). 4. Check positive integer values in order: \(n=1\): \(\frac{1!}{1^1}=1\) \(n=2\): \(\frac{2!}{2^2}=0.5\) \(n=3\): \(\frac{3!}{3^3}=\frac{2}{9}\approx 0.222\) \(n=4\): \(\frac{4!}{4^4}=\frac{3}{32}=0.09375\). The first value below \(0.10\) is \(n=4\).

\(n=4\)

- Use the complement that all birth months are different. - Reduce the number of available unused months for each additional student. - Test consecutive values of \(k\) until the probability crosses \(60\%\).

1. Use the complement that all \(k\) students have different birth months. 2. For \(k\le12\), \(P(\text{all different})=\frac{12}{12}\cdot\frac{11}{12}\cdots\frac{12-k+1}{12}\). 3. Check consecutive values near the threshold: For \(k=4\), \(P(\text{at least one shared month})=1-\frac{12\cdot11\cdot10\cdot9}{12^4}=\frac{41}{96}\approx 0.427\). For \(k=5\), \(P(\text{at least one shared month})=1-\frac{12\cdot11\cdot10\cdot9\cdot8}{12^5}=\frac{89}{144}\approx 0.618\). 4. The probability first exceeds \(0.60\) when \(k=5\).

\(k=5\)

- Consider the caps one at a time and track how many unused designs remain. - Use a complement for “at least two match.” - Think about why relative frequencies from finitely many trials vary.

a) The probability that all six designs are different is \(\frac{25}{25}\cdot\frac{24}{25}\cdot\frac{23}{25}\cdot\frac{22}{25}\cdot\frac{21}{25}\cdot\frac{20}{25}\approx0.52229\). b) The event “at least two match” is the complement, so its probability is \(1-0.52229\approx0.47771\). c) A finite simulation produces a relative frequency that varies randomly from one set of trials to another. With \(1000\) trials, the estimate should often be near the theoretical probability but usually will not equal it exactly.

a) Approximately \(0.52229\), or \(52.23\%\) b) Approximately \(0.47771\), or \(47.77\%\) c) Simulation estimates are subject to random variation.

- First decide whether order matters for an unassigned committee. - In part b, compare one specified committee with all equally likely committees. - Distinct roles make different arrangements of the same three students count separately.

a) Without assigned roles, order does not matter, so the number of committees is \(\binom{20}{3}=1140\). b) Exactly one of the \(1140\) equally likely committees contains the three class representatives. Thus, the probability is \(\frac{1}{1140}\approx0.000877\). c) With three distinct roles, order matters. The number of assignments is \(20\cdot19\cdot18=6840\). Equivalently, \(\binom{20}{3}3!=6840\).

a) \(1140\) committees b) \(\frac{1}{1140}\approx0.000877\), or about \(0.0877\%\) c) \(6840\) role assignments

- For all different results, reduce the number of available new outcomes after each spin. - Use a complement for at least one repeat. - Compare the number of spins with the number of possible results.

1. The probability that all \(4\) results are different is \(\frac{12}{12}\cdot\frac{11}{12}\cdot\frac{10}{12}\cdot\frac{9}{12}=\frac{11880}{20736}\approx 0.57292\). The probability of at least one repeat is the complement, approximately \(0.42708\). Therefore, all different is more likely. 2. There are only \(12\) possible results. By the pigeonhole principle, \(13\) spins must include a repeat, so the probability of all different results is \(0\).

1. All different is more likely: approximately \(0.57292\) versus \(0.42708\). 2. \(0\)

- For all different PINs, track how many unused PINs remain at each assignment. - Use a complement for at least one match. - Think about how the number of possible matching pairs changes as employees are added.

a) Each of the \(10{,}000\) PINs must be equally likely for every employee, and assignments must be independent. b) The probability that all \(50\) PINs differ is \(\frac{10000\cdot9999\cdot9998\cdots9951}{10000^{50}}\approx0.88453\). c) Taking the complement gives \(P(\text{at least one match})=1-0.88453\approx0.11547\). d) With more employees, there are more pairs that could match, and each additional PIN must avoid a larger set of previously assigned PINs.

a) PINs are assigned independently and uniformly from the \(10{,}000\) possibilities. b) Approximately \(0.88453\) c) Approximately \(0.11547\), or \(11.55\%\) d) More employees create more possible matching pairs.

- Represent each assignment as a permutation. - Use the complement of no correct assignments for part b. - Consider what remains after three correct assignments. - Small samples can vary noticeably from theoretical probabilities.

a) The permutations with exactly two correct assignments are \((1,2,4,3)\), \((1,4,3,2)\), \((1,3,2,4)\), \((4,2,3,1)\), \((3,2,1,4)\), and \((2,1,3,4)\). The identity permutation \((1,2,3,4)\) has four correct assignments, so these seven permutations have at least two correct assignments. b) Of the \(4!=24\) permutations, \(9\) have no fixed points. Therefore, \(P(\text{at least one correct})=1-\frac{9}{24}=\frac{5}{8}=0.625\). c) If three guests receive the correct drinks, the only remaining drink must belong to the only remaining guest, so all four assignments are correct. d) The observed relative frequency is \(\frac{15}{20}=0.75\). With only \(20\) trials, a difference from \(0.625\) of this size can result from random variation, so it does not contradict the model.

a) \((1,2,4,3)\), \((1,4,3,2)\), \((1,3,2,4)\), \((4,2,3,1)\), \((3,2,1,4)\), \((2,1,3,4)\), \((1,2,3,4)\) b) \(\frac{5}{8}=0.625\) c) Three correct assignments force the fourth assignment to be correct. d) No. A relative frequency of \(0.75\) in only \(20\) trials can result from random variation.

- Count all ordered triples first. - For part 1, identify the number multisets that sum to \(5\) and count their distinct orders. - A strictly decreasing sequence uses three distinct numbers in one specific order. - For part 3, restrict every draw to the values \(1\) through \(4\).

There are \(10^3=1000\) equally likely ordered triples. 1. The triples with sum \(5\) are the three permutations of \((1,1,3)\) and the three permutations of \((1,2,2)\), for six favorable triples. Thus, \(P=\frac{6}{1000}=\frac{3}{500}\). 2. A strictly decreasing triple is determined by choosing three distinct numbers; each chosen set has exactly one decreasing order. Thus, \(P=\frac{\binom{10}{3}}{1000}=\frac{3}{25}\). 3. If the greatest number is at most \(4\), every draw must be one of \(1,2,3,4\). Thus, \(P=\frac{4^3}{10^3}=\frac{8}{125}\).

1. \(\frac{3}{500}=0.006\) 2. \(\frac{3}{25}=0.12\) 3. \(\frac{8}{125}=0.064\)

- Count all ordered outcomes from four runs. - For part b, first find the probability that every generated value is different. - For part c, count the outcomes in which one value repeats on all four runs.

a) For \(n=6\), there are \(6^4\) ordered outcomes. The number with four different values is \(6\cdot5\cdot4\cdot3\). Thus, \(P=\frac{6\cdot5\cdot4\cdot3}{6^4}=\frac{5}{18}\approx0.2778\). b) The probability that all four numbers are different is \(\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdot\frac{n-3}{n}=\frac{(n-1)(n-2)(n-3)}{n^3}\). Taking the complement gives \(P(\text{at least two equal})=1-\frac{(n-1)(n-2)(n-3)}{n^3}\). c) There are \(n\) ordered outcomes in which all four numbers are the same, out of \(n^4\) total outcomes. Hence, \(\frac{1}{n^3}=0.008\), so \(n^3=125\) and \(n=5\).

a) \(\frac{5}{18}\approx0.2778\) b) \(1-\frac{(n-1)(n-2)(n-3)}{n^3}\) c) \(n=5\)

- For all different cards, track how many choices remain after each distinct result. - Use the complement of “all three are different.” - For the threshold, solve the inequality and check the neighboring integer values.

a) For \(n=10\), \(P(\text{all different})=\frac{10}{10}\cdot\frac{9}{10}\cdot\frac{8}{10}=0.72\). b) For general \(n\), \(P(\text{all different})=\frac{(n-1)(n-2)}{n^2}\). Therefore, \(P(\text{at least two identical})=1-\frac{(n-1)(n-2)}{n^2}=\frac{3n-2}{n^2}\). c) Solve \(\frac{(n-1)(n-2)}{n^2}>0.98\), which is equivalent to \(0.02n^2-3n+2>0\). The relevant boundary is approximately \(149.33\), so the least integer satisfying the inequality is \(n=150\). Checking the neighboring integers confirms that \(n=149\) is below \(98\%\) and \(n=150\) is above it.

a) \(0.72\) b) \(P(\text{at least two identical})=\frac{3n-2}{n^2}\) c) \(n=150\)

- Count the ordered sequences in which every roll gives a new result. - Use a complement for “at least one repeat.” - For the binomial claim, check whether the proposed success probability remains constant across trials.

a) There are \(8^5\) ordered sequences. For all results to be different, there are \(8\cdot7\cdot6\cdot5\cdot4\) valid sequences. Thus, \(P=\frac{105}{512}\approx0.2051\). b) The event is the complement of all results being different, so \(P=1-\frac{105}{512}=\frac{407}{512}\approx0.7949\). c) The claim is false. If “success” means repeating any previous result, its probability changes depending on how many distinct values have already appeared. A binomial model with \(p=\frac{1}{8}\) would instead count occurrences of one specified number.

a) \(\frac{105}{512}\approx0.2051\) b) \(\frac{407}{512}\approx0.7949\) c) The claim is false because the probability of a repeat is not constant from roll to roll.