Free Math Worksheets

- Compare the contents of the box before and after a draw. - Compare the probability of winning on the first draw with the probability on a later draw. - Recall the constant-probability and independence requirements for a binomial experiment.

1. With replacement, each draw has two outcomes, winning or not winning. The probability of a winning ticket remains \(p=\frac{5}{20}=0.25\), and the draws are independent. Therefore, part a can be modeled as a binomial experiment with \(n=4\). 2. Without replacement, the composition of the box changes after each draw. For example, after drawing a winning ticket first, the next winning probability is \(\frac{4}{19}\), not \(\frac{5}{20}\). 3. Therefore, part b is not binomial because the draws are dependent and the success probability is not constant.

a) Yes. Replacing each ticket keeps the winning probability constant at \(0.25\) and makes the draws independent. b) No. Without replacement, the draws are dependent and the winning probability changes from draw to draw.

- Find the initial number of nougat chocolates. - Account for the change in both the total number of chocolates and the number of nougat chocolates after the first selection. - Multiply the two conditional probabilities.

1. There are \(24-4=20\) nougat chocolates. 2. The probability that the first chocolate has nougat filling is \(\frac{20}{24}\). 3. After a nougat chocolate is selected, \(19\) of the remaining \(23\) chocolates have nougat filling. 4. Therefore, \(P(\text{both nougat})=\frac{20}{24}\cdot\frac{19}{23}=\frac{95}{138}\approx0.6884\).

\(\frac{95}{138}\approx 0.6884\)

- Track how the total number of chocolates changes after each selection. - Also track how many chocolates of the desired coating remain. - Multiply the three conditional probabilities for each part.

1. For three dark-chocolate coatings, \(P(\text{three dark})=\frac{8}{12}\cdot\frac{7}{11}\cdot\frac{6}{10}=\frac{14}{55}\). 2. For three milk-chocolate coatings, \(P(\text{three milk})=\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}=\frac{1}{55}\).

a) \(\frac{14}{55}\) b) \(\frac{1}{55}\)

- Update the available coins after the first draw. - For part b, identify both same-color paths. - Multiply along each path and add the two path probabilities.

1. For gold then silver, \(P(\text{gold then silver})=\frac{3}{8}\cdot\frac{5}{7}=\frac{15}{56}\). 2. The same color occurs as gold-gold or silver-silver: \(P(\text{same color})=\frac{3}{8}\cdot\frac{2}{7}+\frac{5}{8}\cdot\frac{4}{7}=\frac{13}{28}\).

a) \(\frac{15}{56}\) b) \(\frac{13}{28}\)

- Compare the second-stage probabilities with the first-stage probabilities. - Decide which procedure leaves the contents of the bag unchanged.

1. In diagram A, the second-draw probabilities remain \(\frac{3}{5}\) and \(\frac{2}{5}\) after either first result. Therefore, diagram A represents drawing with replacement. 2. In diagram B, the second-draw probabilities change and have denominator \(4\). Therefore, diagram B represents drawing without replacement.

A: with replacement B: without replacement

- Follow the one path that satisfies both conditions. - Update the number of aces and the total number of cards after the first draw.

1. The probability that the first card is an ace is \(\frac{2}{8}\). 2. After an ace is drawn, one ace remains among seven cards. 3. Therefore, \(P(\text{two aces})=\frac{2}{8}\cdot\frac{1}{7}=\frac{1}{28}\).

\(\frac{1}{28}\approx 3.57\%\)

- One girl and one boy can be selected in two different orders. - Update the number of students remaining after the first selection.

1. A girl can be selected first and a boy second: \(\frac{4}{6}\cdot\frac{2}{5}=\frac{4}{15}\). 2. A boy can be selected first and a girl second: \(\frac{2}{6}\cdot\frac{4}{5}=\frac{4}{15}\). 3. Add the probabilities of the two possible orders: \(\frac{4}{15}+\frac{4}{15}=\frac{8}{15}\).

\(\frac{8}{15}\approx 53.3\%\)

- Update the numbers of available people after each selection. - List the possible orders for exactly two women. - Use a complement for “at least one woman.” - Different orders may belong to the same event.

1. For three men, \(P=\frac{4}{10}\cdot\frac{3}{9}\cdot\frac{2}{8}=\frac{1}{30}\). 2. Exactly two women can occur in the orders woman-woman-man, woman-man-woman, or man-woman-woman. Each order has probability \(\frac{6\cdot5\cdot4}{10\cdot9\cdot8}=\frac{1}{6}\). Thus, the total probability is \(3\cdot\frac{1}{6}=\frac{1}{2}\). 3. The complement of at least one woman is selecting only men. Therefore, \(P(\text{at least one woman})=1-\frac{1}{30}=\frac{29}{30}\).

1) \(\frac{1}{30}\approx 0.033\) 2) \(\frac{1}{2}=0.5\) 3) \(\frac{29}{30}\approx 0.967\)

- List the possible positions of the one ace. - Remember that the deck size changes after each draw. - Find the probability of one position pattern, then account for all three patterns.

1. The ace can occur on the first, second, or third draw, so there are three favorable position patterns. 2. For the pattern ace, non-ace, non-ace, the probability is \(\frac{4}{52}\cdot\frac{48}{51}\cdot\frac{47}{50}=\frac{376}{5525}\). 3. The other two position patterns have the same probability. Therefore, \(P(\text{exactly one ace})=3\cdot\frac{376}{5525}=\frac{1128}{5525}\approx 0.2042\).

\(\frac{1128}{5525}\approx 0.2042\)

- Determine the number of nonwinning tickets. - Treat the simultaneous selection as drawing without replacement. - Update both the numerator and denominator after each selected ticket. - Multiply the three conditional probabilities.

1. There are \(15-6=9\) nonwinning tickets. Selecting three tickets at once is equivalent to selecting without replacement. 2. For three winning tickets, \(P(\text{three winning})=\frac{6}{15}\cdot\frac{5}{14}\cdot\frac{4}{13}=\frac{4}{91}\). 3. For three nonwinning tickets, \(P(\text{three nonwinning})=\frac{9}{15}\cdot\frac{8}{14}\cdot\frac{7}{13}=\frac{12}{65}\).

a) \(\frac{4}{91}\) b) \(\frac{12}{65}\)

- Recall the defining conditions of a binomial experiment. - Consider when sampling without replacement can be treated as approximately independent. - Compare the sample size with the population size.

1. A binomial approximation is reasonable when each bottle is classified as defective or not defective, the defect probability is approximately constant, and the selections are approximately independent. Sampling from a very large production stream makes the effect of removing one bottle negligible. 2. The sample size gives \(n=100\), and the defective rate gives \(p=0.005\). 3. Sampling \(100\) of only \(120\) bottles without replacement changes the composition substantially after each selection. The selections are strongly dependent, so a binomial model is not appropriate.

a) The model is reasonable when there are two outcomes, the selections are approximately independent, and the defect probability remains approximately constant. b) \(n=100\); \(p=0.005\) c) No. Sampling most of a small population without replacement creates strong dependence and a changing success probability.

- For part a, list the two possible color orders. - Remember that the total number of balls changes after the first draw. - For part b, use the complement of drawing two blue balls.

1. Different colors can occur as yellow-blue or blue-yellow: \(P(\text{different colors})=\frac{4}{10}\cdot\frac{6}{9}+\frac{6}{10}\cdot\frac{4}{9}=\frac{8}{15}\). 2. For part b, use the complement that both balls are blue: \(P(\text{at least one yellow})=1-\frac{6}{10}\cdot\frac{5}{9}=1-\frac{1}{3}=\frac{2}{3}\).

a) \(\frac{8}{15}\) b) \(\frac{2}{3}\)

- Distinguish between selection without replacement and repeated selection with replacement. - Use combinations to count unordered groups. - Divide the number of favorable groups by the total number of groups. - Explain why replacement leads to a binomial model in part b).

1. For part a), the number of possible groups is \(\binom{20}{5}=15{,}504\). 2. A favorable group contains \(3\) of the \(12\) tablet owners and \(2\) of the \(8\) nonowners, so the number of favorable groups is \(\binom{12}{3}\binom{8}{2}=6160\). 3. Therefore, the probability is \(\frac{6160}{15504}=\frac{385}{969}\approx 0.397317\). 4. For part b), replacement makes the selections independent with constant probability \(p=\frac{12}{20}=0.60\). If \(X\sim\operatorname{Bin}(10,0.60)\), then \(P(X=4)=\binom{10}{4}(0.60)^4(0.40)^6=0.111476736\).

a) There are \(15{,}504\) possible groups. The probability is \(\frac{385}{969}\approx 39.73\%\). b) The probability is \(0.111476736\), or approximately \(11.15\%\).

- Distinguish between one group selected without replacement and repeated selections with replacement. - Use combinations to count unordered groups. - In part c), the success probability remains constant over the four selections.

1. For part a), the number of groups is \(\binom{15}{4}=1365\). 2. For part b), the number of favorable groups is \(\binom{6}{2}\binom{9}{2}=540\). Thus, the probability is \(\frac{540}{1365}=\frac{36}{91}\approx 0.395604\). 3. For part c), replacement gives a constant probability \(p=\frac{6}{15}=0.40\). If \(X\sim\operatorname{Bin}(4,0.40)\), then \(P(X=2)=\binom{4}{2}(0.40)^2(0.60)^2=0.3456\).

a) There are \(1365\) possible groups. b) The probability is \(\frac{36}{91}\approx 39.56\%\). c) The probability is \(34.56\%\).

- Count all samples of \(5\) batteries from \(20\). - For an exact count, choose defective and nondefective batteries separately. - “At most \(1\)” includes \(0\) and \(1\). - Use a complement for “at least \(1\).”

1. There are \(\binom{20}{5}\) equally likely samples. 2. For exactly \(2\) defective batteries, \(P(X=2)=\frac{\binom{4}{2}\binom{16}{3}}{\binom{20}{5}}=\frac{3360}{15504}\approx 0.21672\). 3. For at most \(1\) defective battery, \(P(X\le 1)=\frac{\binom{4}{0}\binom{16}{5}+\binom{4}{1}\binom{16}{4}}{\binom{20}{5}}=\frac{11648}{15504}\approx 0.75129\). 4. For at least \(1\) defective battery, use the complement: \(P(X\ge 1)=1-\frac{\binom{16}{5}}{\binom{20}{5}}=\frac{11136}{15504}\approx 0.71827\).

a) \(0.21672\), or about \(21.67\%\) b) \(0.75129\), or about \(75.13\%\) c) \(0.71827\), or about \(71.83\%\)

- Distinguish a sample selected without replacement from repeated selections with replacement. - Use the complement relationship between no defects and at least \(1\) defect. - Consider how removing a nondefective phone changes the next probability.

1. Without replacement, \(P(X=0)=\frac{\binom{21}{4}}{\binom{24}{4}}=\frac{5985}{10626}\approx 0.563241\). 2. For part b), \(P(X\ge 1)=1-P(X=0)\approx 0.436759\). 3. With replacement, the probability of a nondefective phone remains \(\frac{21}{24}=0.875\). Thus, the probability of four nondefective phones is \((0.875)^4\approx 0.586182\). 4. The probability is higher with replacement because selecting a nondefective phone does not reduce the number of nondefective phones available for later selections.

a) The probability is approximately \(56.32\%\). b) The probability is approximately \(43.68\%\). c) With replacement, the probability is approximately \(58.62\%\), which is higher than without replacement.

- Decide whether the selection is made with or without replacement. - Count the total number of ways to select \(4\) students from \(12\). - Translate “more than half” into a condition on \(X\) for a sample of \(4\). - Check that the probabilities in your table add to \(1\).

1. This is sampling without replacement from a population of \(12\) students, with \(7\) students who play an instrument and a sample size of \(4\). For \(k\in\{0,1,2,3,4\}\), \(P(X=k)=\frac{\binom{7}{k}\binom{5}{4-k}}{\binom{12}{4}}\), where \(\binom{12}{4}=495\). 2. The probabilities are \(P(X=0)=\frac{\binom{7}{0}\binom{5}{4}}{495}=\frac{1}{99}\), \(P(X=1)=\frac{\binom{7}{1}\binom{5}{3}}{495}=\frac{14}{99}\), \(P(X=2)=\frac{\binom{7}{2}\binom{5}{2}}{495}=\frac{42}{99}\), \(P(X=3)=\frac{\binom{7}{3}\binom{5}{1}}{495}=\frac{35}{99}\), and \(P(X=4)=\frac{\binom{7}{4}\binom{5}{0}}{495}=\frac{7}{99}\). 3. More than half of \(4\) students means \(X>2\). Therefore, \(P(X>2)=P(X=3)+P(X=4)=\frac{35}{99}+\frac{7}{99}=\frac{14}{33}\approx0.4242\).

a) <table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(\frac{1}{99}\)</td><td>\(\frac{14}{99}\)</td><td>\(\frac{42}{99}\)</td><td>\(\frac{35}{99}\)</td><td>\(\frac{7}{99}\)</td></tr> </table> b) \(P(X>2)=\frac{14}{33}\approx42.42\%\).

- Count samples using combinations for selection without replacement. - Use the complement event of no defective drives. - With replacement, the defect probability remains constant. - The absolute difference is the absolute value of the difference between the two model results.

1. Without replacement, use the complement of selecting \(3\) nondefective drives: \(P(\text{return})=1-\frac{\binom{47}{3}}{\binom{50}{3}}\approx 0.172704\). 2. With replacement, the defect probability is \(p=\frac{3}{50}=0.06\). Thus, \(P(\text{return})=1-(0.94)^3\approx 0.169416\). 3. The absolute difference is \(|0.172704-0.169416|\approx 0.003288\).

a) The probability without replacement is approximately \(17.27\%\). b) The probability with replacement is approximately \(16.94\%\). c) The absolute difference is approximately \(0.0033\), or \(0.33\) percentage points.

- Use a binomial model with replacement and combinations without replacement. - “At least \(2\)” includes exactly \(2\) and exactly \(3\). - Calculate both probabilities before comparing them.

1. With replacement, let \(X\sim\operatorname{Bin}(3,0.60)\). Then \(P(X\ge 2)=\binom{3}{2}(0.60)^2(0.40)+(0.60)^3=0.648\). 2. Without replacement, the number of favorable samples is \(\binom{6}{2}\binom{4}{1}+\binom{6}{3}=60+20=80\), out of \(\binom{10}{3}=120\) total samples. 3. Thus, the probability without replacement is \(\frac{80}{120}=\frac{2}{3}\approx 0.666667\). 4. Since \(0.666667>0.648\), sampling without replacement is more favorable.

Sampling without replacement is more favorable. The win probability is approximately \(66.67\%\), compared with \(64.8\%\) with replacement.

- Find the proportion of batteries that work. - Use a binomial model for independent selections and combinations for selection without replacement. - Compare the sample size with the population size when judging the approximation.

1. With replacement, the probability that a selected battery works is \(p=\frac{45}{50}=0.9\). Therefore, \(P(X=4)=(0.9)^4=0.6561\). 2. Without replacement, \(P(Y=4)=\frac{\binom{45}{4}}{\binom{50}{4}}=\frac{148995}{230300}\approx 0.6470\). 3. The difference is \(0.6561-0.6470\approx 0.0091\), or about \(0.91\) percentage point. Because the sample of \(4\) batteries is less than \(10\%\) of the population of \(50\), the binomial model is a reasonable approximation, though it gives a slightly larger probability.

a) \(0.6561\) b) \(\frac{148995}{230300}\approx 0.6470\) c) The binomial model gives a result about \(0.0091\) higher and is a reasonable approximation because the sample is small relative to the population.

- Count all groups of \(4\) from \(24\). - For a group containing members from two voice parts, multiply the two selection counts. - If two specified people are already included, choose the remaining members from everyone else. - Use one distinct ball per choir member, color the balls by voice part, and mark the two featured soloists.

1. There are \(\binom{24}{4}=10{,}626\) possible quartets. 2. For all sopranos, \(P=\frac{\binom{14}{4}}{\binom{24}{4}}=\frac{1001}{10626}\approx 0.09420\). 3. For two sopranos and two altos, \(P=\frac{\binom{14}{2}\binom{10}{2}}{\binom{24}{4}}=\frac{4095}{10626}\approx 0.38538\). 4. If both specified soloists are selected, the other two members can be any \(2\) of the remaining \(22\). Thus, \(P=\frac{\binom{22}{2}}{\binom{24}{4}}=\frac{231}{10626}\approx 0.02174\). 5. An urn model uses \(24\) distinct balls, one for each choir member. Color \(14\) balls to represent sopranos and \(10\) to represent altos, and specially mark the two balls representing the featured soloists. Draw \(4\) balls without replacement, and ignore order because only the selected group matters.

a) \(\frac{1001}{10626}\approx 0.09420\) b) \(\frac{4095}{10626}\approx 0.38538\) c) \(\frac{231}{10626}\approx 0.02174\) d) Use \(24\) distinct balls, one per member; color \(14\) for sopranos and \(10\) for altos, mark the two featured soloists, and draw \(4\) without replacement while ignoring order.

- Decide whether the number of available copies of a digit changes after each selection. - For Method B, multiply the conditional probability at each draw. - Compare a code with repeated digits with a code containing four different digits.

1. In Method A, the digit selections are independent, so \(P_A(7777)=\left(\frac{1}{10}\right)^4=0.0001\). In Method B, \(P_B(7777)=\frac{4}{40}\cdot\frac{3}{39}\cdot\frac{2}{38}\cdot\frac{1}{37}\approx 0.00001094\). 2. In Method A, \(P_A(1234)=\left(\frac{1}{10}\right)^4=0.0001\). In Method B, \(P_B(1234)=\frac{4}{40}\cdot\frac{4}{39}\cdot\frac{4}{38}\cdot\frac{4}{37}\approx 0.00011672\). 3. Method A is equivalent to sampling with replacement: one digit does not affect the probabilities for later digits. Method B samples without replacement, so each draw changes the composition of the urn. Its draws are dependent, making repeated-digit codes less likely and all-distinct codes more likely than in Method A.

1. Method A: \(0.0001\); Method B: approximately \(0.00001094\) 2. Method A: \(0.0001\); Method B: approximately \(0.00011672\) 3. Method A has independent selections, while Method B has dependent draws because the balls are not replaced. Therefore, the code probabilities are not all equal in Method B.

- Compare the second-draw probability with and without replacement. - Independence requires \(P(A\cap B)=P(A)P(B)\). - Judge the approximation by comparing the sample size with the population size.

1. With replacement, the selections are independent and \(P(\text{defective})=\frac{10}{500}=0.02\). Thus, \(P(\text{both defective})=(0.02)^2=0.0004\). 2. Without replacement, \(P(\text{both defective})=\frac{10}{500}\cdot\frac{9}{499}=\frac{90}{249500}\approx 0.0003607\). 3. By symmetry, \(P(E_1)=P(E_2)=0.02\). However, \(P(E_1\cap E_2)\approx 0.0003607\), while \(P(E_1)P(E_2)=0.0004\). Because these values are unequal, the events are dependent. 4. The sample size of \(2\) is very small compared with the population of \(500\), so removing one item changes the defect probability only slightly. The with-replacement model is therefore a useful approximation.

1. \(0.0004\) 2. \(\frac{90}{249500}\approx 0.0003607\) 3. The events are dependent because \(P(E_1\cap E_2)\ne P(E_1)P(E_2)\). 4. Yes. The sample is very small relative to the population, so the probabilities are close.

- Distinguish one specific card from the group of all aces. - Use combinations for the first \(5\) cards when there is no replacement. - With replacement, use a binomial model because the ace probability stays constant.

1. Each card is equally likely to occupy the fifth position, so \(P(\text{ace of hearts fifth})=\frac{1}{32}=0.03125\). 2. Without replacement, \(P(X=2)=\frac{\binom{4}{2}\binom{28}{3}}{\binom{32}{5}}=\frac{19656}{201376}\approx 0.09761\). 3. With replacement, each draw has ace probability \(p=\frac{4}{32}=\frac{1}{8}\). Therefore, \(P(X=2)=\binom{5}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^3\approx 0.10468\).

a) \(\frac{1}{32}=0.03125\) b) \(\frac{\binom{4}{2}\binom{28}{3}}{\binom{32}{5}}\approx 0.09761\) c) \(\binom{5}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^3\approx 0.10468\)

- Fix the first group and treat its members as the marked members of the population. - Choose overlapping and nonoverlapping members separately.

1. Treat the first group of \(6\) as fixed. The second group is a sample of \(6\) from \(24\), with \(6\) students marked as members of the first group. 2. For exactly \(2\) students in both groups, \(P=\frac{\binom{6}{2}\binom{18}{4}}{\binom{24}{6}}\approx 0.34102\). 3. For no overlap, \(P=\frac{\binom{6}{0}\binom{18}{6}}{\binom{24}{6}}\approx 0.13792\).

a) Approximately \(0.34102\), or \(34.10\%\) b) Approximately \(0.13792\), or \(13.79\%\)

- Choose defective and nondefective chips separately. - Use a complement for “at least one.”

1. For exactly \(2\) defective chips, \(P(X=2)=\frac{\binom{6}{2}\binom{34}{6}}{\binom{40}{8}}\approx 0.26232\). 2. For no defective chips, \(P(X=0)=\frac{\binom{34}{8}}{\binom{40}{8}}\approx 0.23609\). 3. Using the complement, \(P(X\ge 1)=1-P(X=0)\approx 0.76391\).

a) Approximately \(0.26232\), or \(26.23\%\) b) Approximately \(0.23609\), or \(23.61\%\) c) Approximately \(0.76391\), or \(76.39\%\)

- Count all five-laptop samples. - Choose the required numbers from each brand separately. - If Brand C is excluded, select all laptops from the other \(20\).

1. There are \(\binom{25}{5}\) possible samples. 2. For exactly \(2\) from A and \(3\) from B, \(P=\frac{\binom{12}{2}\binom{8}{3}}{\binom{25}{5}}\approx 0.06957\). 3. For no Brand C laptops, all \(5\) come from the \(20\) A-or-B laptops, so \(P=\frac{\binom{20}{5}}{\binom{25}{5}}\approx 0.29181\). 4. For exactly \(2\) from C, choose the other \(3\) from the remaining \(20\), so \(P=\frac{\binom{5}{2}\binom{20}{3}}{\binom{25}{5}}\approx 0.21457\).

a) Approximately \(0.06957\), or \(6.96\%\) b) Approximately \(0.29181\), or \(29.18\%\) c) Approximately \(0.21457\), or \(21.46\%\)

- Identify the possible values of \(X\). - Count selections with exactly \(k\) coupon bags and \(3-k\) noncoupon bags. - Verify that the probabilities add to \(1\).

1. There are \(\binom{15}{3}=455\) equally likely selections of \(3\) bags. For \(k\in\{0,1,2,3\}\), \(P(X=k)=\frac{\binom{6}{k}\binom{9}{3-k}}{\binom{15}{3}}\). 2. \(P(X=0)=\frac{\binom{6}{0}\binom{9}{3}}{455}=\frac{84}{455}\approx0.1846\). 3. \(P(X=1)=\frac{\binom{6}{1}\binom{9}{2}}{455}=\frac{216}{455}\approx0.4747\). 4. \(P(X=2)=\frac{\binom{6}{2}\binom{9}{1}}{455}=\frac{135}{455}\approx0.2967\). 5. \(P(X=3)=\frac{\binom{6}{3}\binom{9}{0}}{455}=\frac{20}{455}\approx0.0440\). 6. The probabilities add to \(1\).

\(P(X=0)\approx18.46\%\), \(P(X=1)\approx47.47\%\), \(P(X=2)\approx29.67\%\), and \(P(X=3)\approx4.40\%\).

- Use combinations to count samples without replacement. - Use a binomial model when each ball is replaced. - Consider how removing a white or black ball changes the next-draw probabilities.

1. Without replacement, \(P(X=2)=\frac{\binom{10}{2}\binom{10}{1}}{\binom{20}{3}}=\frac{450}{1140}=\frac{15}{38}\approx 0.3947\). 2. With replacement, \(p=0.5\), so \(P(X=2)=\binom{3}{2}(0.5)^2(0.5)=0.375\). 3. Without replacement, the draws are negatively dependent: drawing one color makes that color less likely on the next draw. This reduces the likelihood of extreme counts and makes a central count such as \(2\) white balls slightly more likely. With replacement, the draws are independent and the white-ball probability remains \(0.5\).

1. \(\frac{15}{38}\approx 0.3947\) 2. \(0.375\) 3. The without-replacement probability is higher because the dependent draws make central outcomes more likely than they are under the independent with-replacement model.

- First count all ordered outcomes for three rolls. - Once three distinct values are selected, only one ordering is strictly increasing. - For \(X\), separate the cases: all equal, exactly two distinct values, and all different. - Check that the probabilities in the distribution add to \(1\).

1. There are \(12^3=1728\) ordered outcomes. For three different results, there are \(12\cdot11\cdot10=1320\) outcomes, so the probability is \(\frac{1320}{1728}=\frac{55}{72}\approx0.7639\). 2. Choose any \(3\) distinct values from \(12\). Each chosen set has exactly one strictly increasing order, so the probability is \(\frac{\binom{12}{3}}{1728}=\frac{220}{1728}=\frac{55}{432}\approx0.1273\). 3. If \(X=1\), all rolls are equal, giving \(12\) outcomes and \(P(X=1)=\frac{12}{1728}=\frac{1}{144}\). If \(X=2\), choose the repeated value in \(12\) ways, the position of the different value in \(3\) ways, and that different value in \(11\) ways, giving \(396\) outcomes and \(P(X=2)=\frac{11}{48}\). If \(X=3\), all rolls are different, so \(P(X=3)=\frac{55}{72}\).

1. \(P(\text{all different})=\frac{55}{72}\approx0.7639\) 2. \(P(\text{strictly increasing})=\frac{55}{432}\approx0.1273\) 3. <table> <tr><td>\(x\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>\(P(X=x)\)</td><td>\(\frac{1}{144}\)</td><td>\(\frac{11}{48}\)</td><td>\(\frac{55}{72}\)</td></tr> </table>

- Compare the sample size with the population size. - Use the \(5\%\) condition for approximating dependent selections as independent. - Once the probability is treated as constant, use a binomial model.

1. The sampling fraction is \(\frac{50}{25000}=0.002\), or \(0.2\%\), which is well below \(5\%\). Removing one household changes the subscriber proportion only negligibly, so the success probability can be approximated by \(p=0.15\) throughout the sample. 2. Use \(X\sim\operatorname{Bin}(50,0.15)\). Then \(P(X=8)=\binom{50}{8}(0.15)^8(0.85)^{42}\approx 0.14935\).

a) The sample is only \(0.2\%\) of the population, so sampling without replacement changes the probability by a negligible amount. b) \(P(X=8)\approx 0.14935\), or about \(14.93\%\).

- Find \(5\%\) of the population size. - “At most one” includes \(0\) and \(1\) defective chip. - Add the two corresponding binomial probabilities.

1. The condition requires \(n\le 0.05\cdot1200=60\), so the largest allowed sample size is \(60\). 2. Let \(X\sim\operatorname{Bin}(30,0.03)\). Then \(P(X\le 1)=P(X=0)+P(X=1)\). 3. \(P(X=0)=(0.97)^{30}\approx 0.40101\), and \(P(X=1)=\binom{30}{1}(0.03)(0.97)^{29}\approx 0.37207\). 4. Therefore, \(P(X\le 1)\approx 0.77308\).

a) \(n=60\) b) \(P(X\le 1)\approx 0.77308\), or about \(77.31\%\).

- In part a), “more than \(10\)” means \(11\) or \(12\). - In part b), count samples containing \(5\) of the \(17\) long-blooming seedlings. - Ask whether each selection changes the probability for the next selection.

1. For part a), let \(X\sim\operatorname{Bin}(12,0.85)\). Then \(P(X>10)=P(X=11)+P(X=12)=\binom{12}{11}(0.85)^{11}(0.15)+(0.85)^{12}\approx 0.44346\). 2. For part b), sampling is without replacement from a finite set. Thus, \(P(X=5)=\frac{\binom{17}{5}\binom{3}{0}}{\binom{20}{5}}=\frac{6188}{15504}\approx 0.39912\). 3. Part a) treats each seedling as an independent trial with constant success probability \(0.85\), as is reasonable for purchases from a large supply. In part b), the sample is a substantial fraction of a small, fixed group, so each selection changes the probabilities and a binomial model is not appropriate.

a) \(P(X>10)\approx 0.44346\), or about \(44.35\%\) b) \(\frac{6188}{15504}\approx 0.39912\), or about \(39.91\%\) c) Part a) uses independent trials with a constant probability; part b) uses dependent sampling without replacement from a small finite group.

- Note that the balls are not replaced. - Identify the possible values of \(X\). - Count selections with exactly \(k\) target balls and \(3-k\) other balls. - Use the sample size times the target proportion for the expected value.

1. This is sampling without replacement from \(20\) balls, with \(5\) target balls and a sample size of \(3\). For \(k\in\{0,1,2,3\}\), \(P(X=k)=\frac{\binom{5}{k}\binom{15}{3-k}}{\binom{20}{3}}\), where \(\binom{20}{3}=1140\). 2. The probabilities are \(P(X=0)=\frac{91}{228}\approx0.3991\), \(P(X=1)=\frac{35}{76}\approx0.4605\), \(P(X=2)=\frac{5}{38}\approx0.1316\), and \(P(X=3)=\frac{1}{114}\approx0.0088\). 3. The expected value is \(E(X)=3\cdot\frac{5}{20}=0.75\).

<table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td></tr> <tr><td>\(P(X=k)\)</td><td>\(\frac{91}{228}\approx0.3991\)</td><td>\(\frac{35}{76}\approx0.4605\)</td><td>\(\frac{5}{38}\approx0.1316\)</td><td>\(\frac{1}{114}\approx0.0088\)</td></tr> </table> \(E(X)=0.75\)

- Count the total ways to select \(4\) students from \(24\). - For each \(k\), count groups with \(k\) tablet users and \(4-k\) students without tablets. - Use combinations because order does not matter. - Relate the expected count to the sample size and population proportion.

1. This is sampling without replacement with population size \(24\), \(8\) tablet users, and sample size \(4\). For \(k\in\{0,1,2,3,4\}\), \(P(Y=k)=\frac{\binom{8}{k}\binom{16}{4-k}}{\binom{24}{4}}\). 2. The probabilities are \(P(Y=0)=\frac{130}{759}\approx0.1713\), \(P(Y=1)=\frac{320}{759}\approx0.4216\), \(P(Y=2)=\frac{240}{759}\approx0.3162\), \(P(Y=3)=\frac{64}{759}\approx0.0843\), and \(P(Y=4)=\frac{5}{759}\approx0.0066\). 3. The expected value is \(E(Y)=4\cdot\frac{8}{24}=\frac{4}{3}\approx1.33\).

<table> <tr><td>\(k\)</td><td>\(0\)</td><td>\(1\)</td><td>\(2\)</td><td>\(3\)</td><td>\(4\)</td></tr> <tr><td>\(P(Y=k)\)</td><td>\(\frac{130}{759}\)</td><td>\(\frac{320}{759}\)</td><td>\(\frac{240}{759}\)</td><td>\(\frac{64}{759}\)</td><td>\(\frac{5}{759}\)</td></tr> </table> \(E(Y)=\frac{4}{3}\approx1.33\)

- Update the total number of balls after the first draw. - For part a, identify all three same-color paths. - Multiply along each path and add the favorable paths.

1. The same-color event consists of green-green, blue-blue, and red-red: \(P(\text{same color})=\frac{4}{10}\cdot\frac{3}{9}+\frac{3}{10}\cdot\frac{2}{9}+\frac{3}{10}\cdot\frac{2}{9}=\frac{4}{15}\). 2. For green then red, \(P(\text{green then red})=\frac{4}{10}\cdot\frac{3}{9}=\frac{2}{15}\).

a) \(\frac{4}{15}\) b) \(\frac{2}{15}\)

- Write the red-red and blue-blue paths for each sampling method. - Without replacement, update both the number of balls of the selected color and the total. - For part c, use the complement of drawing two blue balls.

1. With replacement, \(P(\text{same color})=\left(\frac{4}{10}\right)^2+\left(\frac{6}{10}\right)^2=\frac{13}{25}=0.52\). 2. Without replacement, \(P(\text{same color})=\frac{4}{10}\cdot\frac{3}{9}+\frac{6}{10}\cdot\frac{5}{9}=\frac{7}{15}\). 3. Use the complement of drawing no red balls. With replacement: \(P(\text{at least one red})=1-\left(\frac{6}{10}\right)^2=\frac{16}{25}=0.64\). Without replacement: \(P(\text{at least one red})=1-\frac{6}{10}\cdot\frac{5}{9}=\frac{2}{3}\). Since \(\frac{2}{3}>\frac{16}{25}\), the probability is greater without replacement.

a) \(\frac{13}{25}=52\%\) b) \(\frac{7}{15}\approx 46.7\%\) c) Without replacement: \(\frac{2}{3}\approx 66.7\%\), compared with \(\frac{16}{25}=64\%\) with replacement.

- Update the number of balls after the first draw. - A same-color pair is possible only for colors with at least two balls. - For part b, use the complement of drawing two nonred balls.

1. The same-color outcomes are blue-blue and red-red. Therefore, \(P(\text{same color})=\frac{3}{6}\cdot\frac{2}{5}+\frac{2}{6}\cdot\frac{1}{5}=\frac{8}{30}=\frac{4}{15}\). 2. For part b, use the complement of drawing no red balls. There are \(4\) nonred balls, so \(P(\text{no red})=\frac{4}{6}\cdot\frac{3}{5}=\frac{2}{5}\). 3. Thus, \(P(\text{at least one red})=1-\frac{2}{5}=\frac{3}{5}\).

a) \(\frac{4}{15}\approx 26.7\%\) b) \(\frac{3}{5}=60\%\)

- Count the balls of each color. - Update the counts after the first draw. - For part b, calculate the complement of drawing two balls of the same color.

1. For two red balls, \(P(\text{red, red})=\frac{4}{9}\cdot\frac{3}{8}=\frac{1}{6}\). 2. Use the complement that the balls have the same color: \(P(\text{same color})=\frac{4}{9}\cdot\frac{3}{8}+\frac{3}{9}\cdot\frac{2}{8}+\frac{2}{9}\cdot\frac{1}{8}=\frac{5}{18}\). Therefore, \(P(\text{different colors})=1-\frac{5}{18}=\frac{13}{18}\).

a) \(\frac{1}{6}\) b) \(\frac{13}{18}\)

- Count the balls showing each letter. - Determine whether the probabilities remain constant after each draw. - For part c, count the possible orders of N, A, and S.

1. With replacement, \(P(\text{NAS})=\frac{2}{6}\cdot\frac{3}{6}\cdot\frac{1}{6}=\frac{1}{36}\). 2. Without replacement, \(P(\text{NAS})=\frac{2}{6}\cdot\frac{3}{5}\cdot\frac{1}{4}=\frac{1}{20}\). 3. To rearrange the letters into NAS, the sample must contain one N, one A, and one S. These letters have \(3!=6\) possible orders, each with probability \(\frac{1}{20}\). Therefore, \(P(\text{one each of N, A, and S})=6\cdot\frac{1}{20}=\frac{3}{10}\).

a) \(\frac{1}{36}\) b) \(\frac{1}{20}\) c) \(\frac{3}{10}\)

- Count all balls by color. - Because the draws are without replacement, the number of available balls changes after the first draw. - List every possible color combination and its total gain or loss. - Add the probabilities of combinations that produce the same value of \(X\). - Organize the values and probabilities in a table.

1. There are \(9\cdot8=72\) equally likely ordered draws when the physical balls are distinguished. 2. The possible totals and ordered-path counts are: \(-6\) from two yellow balls, \(2\) paths; \(-2\) from one red and one yellow, \(12\) paths; \(-1\) from one blue and one yellow, \(16\) paths; \(2\) from two red balls, \(6\) paths; \(3\) from one blue and one red, \(24\) paths; and \(4\) from two blue balls, \(12\) paths. 3. Dividing these counts by \(72\) gives the probability distribution.

<table border="1"> <thead> <tr><th>\(k\), in dollars</th><th>\(-6\)</th><th>\(-2\)</th><th>\(-1\)</th><th>\(2\)</th><th>\(3\)</th><th>\(4\)</th></tr> </thead> <tbody> <tr><th>\(P(X=k)\)</th><td>\(\frac{1}{36}\)</td><td>\(\frac{1}{6}\)</td><td>\(\frac{2}{9}\)</td><td>\(\frac{1}{12}\)</td><td>\(\frac{1}{3}\)</td><td>\(\frac{1}{6}\)</td></tr> </tbody> </table>

- List the two color paths that end with green. - Update the bag’s composition after the first draw. - Multiply along each path and add the two path probabilities.

1. The second ball can be green after either a green first draw or an orange first draw. 2. For green then green, \(P(GG)=\frac{3}{5}\cdot\frac{2}{4}=\frac{3}{10}\). 3. For orange then green, \(P(OG)=\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{10}\). 4. Therefore, \(P(\text{second ball green})=\frac{3}{10}+\frac{3}{10}=\frac{3}{5}\).

\(\frac{3}{5}=0.6\)

- Update the counts after the first draw. - For part a, add the three same-color paths. - For part b, treat orange and gray together as not purple. - For part c, use the complement of drawing no orange balls.

1. For the same color, add the orange-orange, purple-purple, and gray-gray paths: \(P(\text{same color})=\frac{2}{10}\cdot\frac{1}{9}+\frac{3}{10}\cdot\frac{2}{9}+\frac{5}{10}\cdot\frac{4}{9}=\frac{14}{45}\). 2. Seven balls are not purple, so \(P(\text{no purple})=\frac{7}{10}\cdot\frac{6}{9}=\frac{7}{15}\). 3. Use the complement of no orange. Eight balls are not orange, so \(P(\text{at least one orange})=1-\frac{8}{10}\cdot\frac{7}{9}=\frac{17}{45}\).

a) \(\frac{14}{45}\) b) \(\frac{7}{15}\) c) \(\frac{17}{45}\)

- List the yellow-green and green-yellow paths. - Calculate the two sampling methods separately. - Without replacement, update the counts before the second draw.

1. With replacement, different colors can occur as yellow-green or green-yellow: \(P(\text{different, with replacement})=\frac{3}{5}\cdot\frac{2}{5}+\frac{2}{5}\cdot\frac{3}{5}=\frac{12}{25}=0.48\). 2. Without replacement, \(P(\text{different, without replacement})=\frac{3}{5}\cdot\frac{2}{4}+\frac{2}{5}\cdot\frac{3}{4}=\frac{3}{5}=0.60\). 3. Since \(0.60>0.48\), two different colors are more likely without replacement.

Without replacement: \(\frac{3}{5}=60\%\), compared with \(\frac{12}{25}=48\%\) with replacement.

- Consider the two possible orders for each pair of different numbers. - Equality cannot occur. - Use symmetry between increasing and decreasing ordered pairs.

1. There are \(5\cdot4=20\) equally likely ordered pairs of different numbers. 2. For every unordered pair of numbers, exactly one order has the second number greater than the first, and the other order has the first number greater than the second. 3. Equality is impossible without replacement, so the two cases are equally likely. Therefore, \(P(\text{second greater than first})=\frac{1}{2}\).

\(\frac{1}{2}=50\%\)

- Count the total number of balls. - Count unordered pairs because the payout depends only on the colors drawn. - Find the favorable color combinations for each positive payout. - Use a complement to find the probability of drawing at least one gray ball.

1. There are \(\binom{12}{2}=66\) equally likely unordered selections. 2. Two purple balls occur in \(\binom{2}{2}=1\) selection, so \(P(X=15)=\frac{1}{66}\). 3. Two orange balls occur in \(\binom{4}{2}=6\) selections, so \(P(X=5)=\frac{1}{11}\). 4. One purple and one orange ball occur in \(2\cdot4=8\) selections, so \(P(X=2)=\frac{4}{33}\). 5. At least one gray ball is the complement of drawing two nongray balls. Thus, \(P(X=-1)=1-\frac{\binom{6}{2}}{66}=\frac{17}{22}\).

\(P(X=15)=\frac{1}{66}\), \(P(X=5)=\frac{1}{11}\), \(P(X=2)=\frac{4}{33}\), and \(P(X=-1)=\frac{17}{22}\).

- Update the counts after the first draw. - For part b, add the two possible color orders. - For part c, use the complement from part a.

1. For two yellow balls, \(P(YY)=\frac{5}{8}\cdot\frac{4}{7}=\frac{5}{14}\). 2. Different colors can occur as yellow-green or green-yellow: \(P(\text{different colors})=\frac{5}{8}\cdot\frac{3}{7}+\frac{3}{8}\cdot\frac{5}{7}=\frac{15}{28}\). 3. Use the complement of two yellow balls: \(P(\text{at least one green})=1-\frac{5}{14}=\frac{9}{14}\).

a) \(\frac{5}{14}\) b) \(\frac{15}{28}\) c) \(\frac{9}{14}\)

- Count how many balls show each letter. - Check whether each draw uses replacement. - For a specified sequence, multiply the probabilities in order. - For part c, count the possible orders of A, K, and T.

1. The letter counts are two balls labeled S, three labeled T, one labeled A, two labeled I, and one labeled K. 2. With replacement, the probabilities stay constant. Therefore, \(P(\text{SIS})=\frac{2}{9}\cdot\frac{2}{9}\cdot\frac{2}{9}=\frac{8}{729}\). 3. Without replacement, \(P(\text{TIT})=\frac{3}{9}\cdot\frac{2}{8}\cdot\frac{2}{7}=\frac{1}{42}\). 4. The letters A, K, and T can be arranged in \(3!=6\) orders. Each order has probability \(\frac{1}{9}\cdot\frac{1}{8}\cdot\frac{3}{7}=\frac{1}{168}\). Thus, \(P(\text{A, K, and T in any order})=6\cdot\frac{1}{168}=\frac{1}{28}\).

a) \(\frac{8}{729}\) b) \(\frac{1}{42}\) c) \(\frac{1}{28}\)

- List the possible positions of the one blue ball. - Update the fractions after each draw. - Add the probabilities of the favorable paths.

1. The favorable color sequences are RRB, RBR, and BRR. 2. Their probabilities are \(P(RRB)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{5}\), \(P(RBR)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{5}\), and \(P(BRR)=\frac{2}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}=\frac{1}{5}\). 3. Therefore, \(P(\text{exactly two red})=\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}\).

\(\frac{3}{5}=60\%\)

- Identify one favorable path for each color. - Multiply along each complete path, then add the three probabilities.

1. The same-color paths are red-red, green-green, and blue-blue. 2. Their total probability is \(\frac{4}{9}\cdot\frac{3}{8}+\frac{3}{9}\cdot\frac{2}{8}+\frac{2}{9}\cdot\frac{1}{8}=\frac{20}{72}=\frac{5}{18}\).

\(\frac{5}{18}\approx 27.8\%\)

- In each model, use the complement of drawing only gray balls. - Compare the second gray branch in the two diagrams.

1. With replacement, use the complement of two gray draws: \(P(\text{at least one gold, with replacement})=1-\left(\frac{4}{5}\right)^2=\frac{9}{25}=0.36\). 2. Without replacement, \(P(\text{at least one gold, without replacement})=1-\frac{4}{5}\cdot\frac{3}{4}=\frac{2}{5}=0.40\). 3. The probability is \(0.04\) greater without replacement.

With replacement: \(0.36\). Without replacement: \(0.40\). Without replacement is greater by \(0.04\).

- Determine the defect probability for one selection. - Use the binomial formula for an exact count with replacement. - Use the complement of no defects for “at least \(1\).” - In part c), track how the conditional probability of selecting a nondefective component changes after one is removed.

1. With replacement, the defect probability remains \(p=\frac{30}{150}=0.20\), so \(X\sim\operatorname{Bin}(10,0.20)\). 2. For part a), \(P(X=2)=\binom{10}{2}(0.20)^2(0.80)^8\approx 0.301990\). 3. For part b), \(P(X\ge 1)=1-(0.80)^{10}\approx 0.892626\). 4. Without replacement, after a nondefective component is selected, the next conditional probability of another nondefective component is slightly less than \(0.80\). Thus, the probability of selecting only nondefective components is less without replacement. Equivalently, \(\frac{120}{150}\cdot\frac{119}{149}\cdots\frac{111}{141}<(0.80)^{10}\).

a) The probability is approximately \(30.20\%\). b) The probability is approximately \(89.26\%\). c) The probability of selecting no defective components is less without replacement than with replacement.

- Decide whether the composition remains constant after each draw. - Match each part to sampling with replacement or without replacement. - For the minimum number of draws, use the complement of no winning tickets. - Reverse the inequality when dividing by a negative logarithm.

1. For part a), let \(X\sim\operatorname{Bin}(6,\frac{1}{3})\). Then \(P(X=2)=\binom{6}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4=\frac{80}{243}\approx 0.329218\). 2. For part b), the probability is \(\frac{\binom{5}{2}\binom{10}{4}}{\binom{15}{6}}=\frac{420}{1001}\approx 0.419580\). 3. For part c), require \(1-\left(\frac{2}{3}\right)^n>0.95\), or \(\left(\frac{2}{3}\right)^n<0.05\). 4. Solving gives \(n>\frac{\ln(0.05)}{\ln(2/3)}\approx 7.388\), so the minimum is \(8\).

a) \(\frac{80}{243}\approx 32.92\%\) b) \(\frac{420}{1001}\approx 41.96\%\) c) At least \(8\) draws are needed.

- Use a binomial model when the success probability remains constant. - Use combinations for sampling without replacement. - Divide the absolute difference by the reference probability to find relative difference. - Compare how much one removal changes the color proportions in each urn.

1. With replacement, both urns have \(p=\frac{2}{3}\). Thus, \(P(X=2)=\binom{3}{2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)=\frac{4}{9}\approx 0.444444\). 2. Without replacement for Urn A, \(P(Y_A=2)=\frac{\binom{8}{2}\binom{4}{1}}{\binom{12}{3}}=\frac{28}{55}\approx 0.509091\). 3. Without replacement for Urn B, \(P(Y_B=2)=\frac{\binom{80}{2}\binom{40}{1}}{\binom{120}{3}}=\frac{3160}{7021}\approx 0.450078\). 4. The relative difference for Urn A is \(\frac{|28/55-4/9|}{4/9}=\frac{8}{55}\approx 14.55\%\). 5. The relative difference for Urn B is \(\frac{|3160/7021-4/9|}{4/9}=\frac{89}{7021}\approx 1.27\%\). 6. For the same sample size and color ratio, removing one ball changes the proportions less in a larger population. Therefore, sampling without replacement more closely resembles independent sampling with replacement in Urn B.

a) Urn A and Urn B: \(\frac{4}{9}\approx 0.4444\) b) Urn A: \(\frac{28}{55}\approx 0.5091\) Urn B: \(\frac{3160}{7021}\approx 0.4501\) c) Urn A: approximately \(14.55\%\) Urn B: approximately \(1.27\%\) d) With a larger population and the same sample size, sampling without replacement is closer to sampling with replacement.

- Determine the defect probability in each shipment. - Count samples with exactly \(1\) defective battery and divide by the total number of samples. - Compare each exact result with the binomial approximation. - Evaluate the ratio of sample size to population size.

1. Both shipments have defect proportion \(p=0.10\). The binomial approximation is \(\binom{4}{1}(0.10)(0.90)^3=0.2916\). 2. For Shipment 1, the exact probability is \(\frac{\binom{5}{1}\binom{45}{3}}{\binom{50}{4}}\approx 0.308076\). 3. For Shipment 2, the exact probability is \(\frac{\binom{50}{1}\binom{450}{3}}{\binom{500}{4}}\approx 0.293164\). 4. For Shipment 1, \(4>0.05\cdot50=2.5\), so the guideline is not met, and the approximation differs noticeably from the exact value. 5. For Shipment 2, \(4\le 0.05\cdot500=25\), so the guideline is met, and the approximation is close to the exact value.

a) The binomial approximation is \(0.2916\). b) Shipment 1: approximately \(0.3081\) Shipment 2: approximately \(0.2932\) c) The guideline is not met for Shipment 1 and is met for Shipment 2. The approximation is much better for Shipment 2.

- How many unordered selections of \(3\) balls can be made from \(10\)? - For “at least one,” consider the complement. - When can two event probabilities be added directly? - Can both specified balls appear in the same selection?

1. There are \(\binom{10}{3}=120\) equally likely selections. If ball \(1\) is included, the other two balls can be chosen in \(\binom{9}{2}=36\) ways. Thus the probability is \(\frac{36}{120}=\frac{3}{10}=0.3\). 2. For the complement, select all three balls from the other eight: \(\binom{8}{3}=56\) selections. Therefore, the probability of selecting at least one of balls \(1\) and \(2\) is \(1-\frac{56}{120}=\frac{8}{15}\approx 0.533\). 3. The claim is false because \(\frac{8}{15}\ne 0.6\). The events “ball \(1\) is selected” and “ball \(2\) is selected” overlap: both balls can be selected. Adding their probabilities without subtracting the overlap counts those selections twice.

a) \(0.3\) b) \(\frac{8}{15}\approx 0.533\) c) The claim is false. The two events are not disjoint, so adding their probabilities double-counts selections containing both balls.

- Let a variable represent the number of red balls. - Express the red-then-blue path probability using that variable. - Solve the resulting equation and keep integer solutions between \(0\) and \(8\).

1. Let \(r\) be the number of red balls. Then \(8-r\) is the number of blue balls. 2. The red-then-blue path gives \(\frac{r}{8}\cdot\frac{8-r}{7}=\frac{15}{56}\). 3. Therefore, \(r(8-r)=15\), so \(r^2-8r+15=0\), which factors as \((r-3)(r-5)=0\). 4. Thus, \(r=3\) or \(r=5\). The bag contains either three red and five blue balls or five red and three blue balls.

Either \(3\) red and \(5\) blue balls, or \(5\) red and \(3\) blue balls.

- Write the red-red path probability in terms of \(x\). - After one red ball is drawn, one fewer red ball remains. - Solve the equation and reject values that cannot be a ball count.

1. The probability of two red balls is \(\frac{x}{10}\cdot\frac{x-1}{9}=\frac{2}{15}\). 2. Multiply and simplify: \(x(x-1)=12\), so \(x^2-x-12=0\). 3. Factoring gives \((x-4)(x+3)=0\). The only valid number of balls is \(x=4\).

\(x=4\)