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Sampling with and without replacement

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Difficulty: 1 – 5

55144312

Use the urn shown. One ball is selected at random. What is the probability that the selected ball is blue?

Figure for problem 551443

Hints

- Count the blue balls shown in the urn. - Compare that count with the total number of balls shown.

Solution

1. The urn shows \(2\) blue balls out of \(5\) balls total. 2. Therefore, the probability is \(\frac{2}{5}\).

Answer

\(\frac{2}{5}\)

55144412

A bag contains 4 green balls and 1 yellow ball. One ball is drawn, its color is recorded, and the ball is returned to the bag before a second draw. What is the probability that the second ball is yellow?

Hints

- Focus on what the bag contains immediately before the second draw. - Returning the first ball restores the original composition.

Solution

1. Because the first ball is returned, the bag has its original composition before the second draw. 2. One of the \(5\) balls is yellow, so the probability is \(\frac{1}{5}\).

Answer

\(\frac{1}{5}\)

55584712

Use the urn shown. One B ball is drawn and not replaced. What is the probability that the next ball drawn is R?

Figure for problem 555847

Hints

- Read the initial R and B counts from the urn. - Update the composition after the stated B ball is removed. - Use the remaining total, not the original total, for the next draw.

Solution

1. The urn initially shows \(3\) R balls and \(2\) B balls. 2. After one B ball is removed, \(3\) R balls and \(1\) B ball remain, for \(4\) balls total. 3. Therefore, the probability that the next ball is R is \(\frac{3}{4}\).

Answer

\(\frac{3}{4}\)

55584912

A bag contains equal numbers of green and orange balls. A green ball is drawn first and is not replaced. Does knowing that the first ball was green change the probability that the second ball is green? Answer yes or no and explain why.

Hints

- Focus on the composition of the bag after the first draw. - Ask whether the proportion of green balls is still the same after a green ball is removed.

Solution

1. Before the first draw, half of the balls are green. 2. A green ball is removed and not returned, so the remaining bag has a smaller proportion of green balls than before. 3. Therefore, knowing the first result changes the probability of green on the second draw.

Answer

Yes. Removing a green ball without replacement lowers the proportion of green balls remaining, so the second-draw probability changes.

52334812

A box contains \(20\) tickets, including \(5\) winning tickets. Four tickets are drawn in sequence. Determine whether each procedure can be modeled as a binomial experiment. a) Each ticket is replaced before the next draw. b) Each ticket is not replaced before the next draw. Briefly justify each answer.

Hints

- 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.

Solution

a) Yes. With replacement, each draw has two relevant outcomes, the winning probability remains \(p=\frac{5}{20}=0.25\), and the draws are independent. The procedure is binomial with \(n=4\). b) No. Without replacement, the composition changes after each draw. For example, after drawing a winning ticket first, the next winning probability is \(\frac{4}{19}\), not \(\frac{5}{20}\). The draws are dependent and the success probability is not constant.

Answer

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.

52338912

A box contains \(24\) chocolates. Four have marzipan filling, and the rest have nougat filling. Two chocolates are selected at random without replacement. Find the probability that both chocolates have nougat filling.

Hints

- 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.

Solution

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, \(P(\text{both nougat})=\frac{20}{24}\cdot\frac{19}{23}=\frac{95}{138}\approx0.6884\).

Answer

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

52339712

A box contains \(12\) chocolates: \(8\) are coated in dark chocolate and \(4\) are coated in milk chocolate. Three chocolates are selected without replacement. a) Find the probability that all three are coated in dark chocolate. b) Find the probability that all three are coated in milk chocolate.

Hints

- 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.

Solution

a) \(P(\text{three dark})=\frac{8}{12}\cdot\frac{7}{11}\cdot\frac{6}{10}=\frac{14}{55}\). b) \(P(\text{three milk})=\frac{4}{12}\cdot\frac{3}{11}\cdot\frac{2}{10}=\frac{1}{55}\).

Answer

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

52381612

A bag contains three gold-colored coins and five silver-colored coins. Two coins are drawn without replacement. a) Find the probability of drawing a gold-colored coin first and a silver-colored coin second. b) Find the probability that both coins have the same color.

Hints

- 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.

Solution

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

Answer

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

53202312

Use the urn shown. Two balls are drawn without replacement. a) Find the probability that the two balls have the same color. b) Find the probability that the first ball is green and the second ball is red. Give each answer as a simplified fraction.

Figure for problem 532023

Hints

- Read the color counts from the urn. - For part a, identify all possible same-color paths. - Update the total number of balls after the first draw.

Solution

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

Answer

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

53611512

Use the urn shown. Two balls are drawn without replacement. Find the probability that the second ball is green.

Figure for problem 536115

Hints

- Read the color counts from the urn. - List the two color paths that end with green. - Update the urn composition after the first draw.

Solution

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

Answer

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

53752512

Use the probability tree shown for three draws without replacement. Find the probability of drawing exactly two red balls.

Figure for problem 537525

Hints

- Trace every path that ends with exactly two R outcomes. - Multiply the probabilities along each complete favorable path. - Add the probabilities of the mutually exclusive favorable paths.

Solution

1. The favorable paths are RRB, RBR, and BRR. 2. From the tree, \(P(RRB)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{5}\). 3. Similarly, \(P(RBR)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{5}\) and \(P(BRR)=\frac{2}{5}\cdot\frac{3}{4}\cdot\frac{2}{3}=\frac{1}{5}\). 4. Add the favorable path probabilities: \(\frac{1}{5}+\frac{1}{5}+\frac{1}{5}=\frac{3}{5}\).

Answer

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

53752912

Two tree diagrams represent two draws from a bag containing \(3\) red balls and \(2\) blue balls. Match each diagram with “with replacement” or “without replacement,” and justify your choices.

Figure for problem 537529

Hints

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

Solution

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.

Answer

A: with replacement B: without replacement

53753212

Use the probability tree shown for two card draws without replacement. Find the probability that both cards are aces.

Figure for problem 537532

Hints

- Identify the single path that satisfies both conditions. - Multiply the probabilities written along that path.

Solution

1. Follow the ace-then-ace path in the tree. 2. Its probability is \(\frac{2}{8}\cdot\frac{1}{7}=\frac{1}{28}\).

Answer

\(\frac{1}{28}\approx3.57\%\)

53753312

Use the probability tree shown for two draws without replacement. Find the probability that the two balls have the same color.

Figure for problem 537533

Hints

- Identify one favorable complete path for each color. - Multiply along each favorable path. - Add the probabilities of the mutually exclusive favorable paths.

Solution

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

Answer

\(\frac{5}{18}\approx27.8\%\)

53755712

Use the probability tree shown for two selections without replacement. Find the probability that one girl and one boy are selected.

Figure for problem 537557

Hints

- One girl and one boy can occur in two different orders. - Multiply along each favorable path in the tree. - Add the probabilities of those mutually exclusive paths.

Solution

1. The favorable paths are girl-then-boy and boy-then-girl. 2. From the tree, \(P(GB)=\frac{4}{6}\cdot\frac{2}{5}=\frac{4}{15}\). 3. Also, \(P(BG)=\frac{2}{6}\cdot\frac{4}{5}=\frac{4}{15}\). 4. Add the two favorable path probabilities: \(\frac{4}{15}+\frac{4}{15}=\frac{8}{15}\).

Answer

\(\frac{8}{15}\approx53.3\%\)

52184712

A work group has \(6\) women and \(4\) men. Three people are selected at random without replacement for a presentation. Find the probability that: 1) Only men are selected. 2) Exactly two women are selected. 3) At least one woman is selected.

Hints

- 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.

Solution

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

Answer

1) \(\frac{1}{30}\approx0.033\) 2) \(\frac{1}{2}=0.5\) 3) \(\frac{29}{30}\approx0.967\)

52339012

A standard \(52\)-card deck contains exactly four aces. Three cards are drawn without replacement. Find the probability that exactly one of the three cards is an ace.

Hints

- 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.

Solution

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, \(P(\text{exactly one ace})=3\cdot\frac{376}{5525}=\frac{1128}{5525}\approx 0.2042\).

Answer

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

52340612

A crate contains \(20\) apples, and \(4\) of them are bruised. Three apples are selected at random without replacement. Find the probability of each event. a) All three apples are bruised. b) None of the three apples is bruised. c) One specified apple, the largest apple in the crate, is selected. d) Exactly two of the selected apples are bruised.

Hints

- Treat the three apples as an unordered sample without replacement. - Count favorable samples and compare them with all three-apple samples. - For an exact bruised/unbruised composition, choose from the two categories separately.

Solution

a) There are \(\binom{20}{3}\) equally likely three-apple samples. Thus, \(P(\text{all bruised})=\frac{\binom{4}{3}}{\binom{20}{3}}=\frac{1}{285}\). b) There are \(16\) unbruised apples, so \(P(\text{none bruised})=\frac{\binom{16}{3}}{\binom{20}{3}}=\frac{28}{57}\). c) If the specified apple is included, choose two more apples from the remaining \(19\). Thus, \(P=\frac{\binom{19}{2}}{\binom{20}{3}}=\frac{3}{20}\). d) Choose two of the four bruised apples and one of the sixteen unbruised apples. Thus, \(P=\frac{\binom{4}{2}\binom{16}{1}}{\binom{20}{3}}=\frac{8}{95}\).

Answer

a) \(\frac{1}{285}\) b) \(\frac{28}{57}\) c) \(\frac{3}{20}\) d) \(\frac{8}{95}\)

52357612

A large bottling plant inspects glass bottles for cracks. Historically, \(0.5\%\) of the bottles are defective. A sample of \(100\) bottles is selected from ongoing production. a) Explain the conditions under which this situation can be approximated by a binomial experiment. b) If finding a defective bottle is a success, state \(n\) and \(p\). c) Suppose instead that a crate contains only \(120\) bottles and \(100\) are inspected without replacement. Discuss whether a binomial model is still appropriate.

Hints

- 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.

Solution

a) 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. b) The sample size gives \(n=100\), and the defective rate gives \(p=0.005\). c) 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.

Answer

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.

52381512

A bag contains four yellow balls and six blue balls. Two balls are drawn without replacement. a) Find the probability that the two balls have different colors. b) Find the probability that at least one ball is yellow.

Hints

- 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.

Solution

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

Answer

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

52688412

A class has \(20\) students. Twelve own a tablet and \(8\) do not. a) Five students are selected at random without replacement. Use combinations to find the total number of possible groups. Then find the probability that exactly \(3\) selected students own a tablet. b) In a different study, one student is selected at random, surveyed, and returned to the group. This procedure is repeated \(10\) times. Find the probability that a student who owns a tablet is selected exactly \(4\) times.

Hints

- Distinguish between selection without replacement and repeated selection with replacement. - Use combinations to count unordered groups. - Explain why replacement leads to a binomial model in part b).

Solution

1. Without replacement, there are \(\binom{20}{5}=15{,}504\) possible groups. A favorable group contains \(3\) of the \(12\) tablet owners and \(2\) of the \(8\) nonowners, giving \(\binom{12}{3}\binom{8}{2}=6160\) groups. Thus the probability is \(\frac{6160}{15{,}504}=\frac{385}{969}\approx0.397317\). 2. With replacement, the selections are independent with constant probability \(p=\frac{12}{20}=0.60\). If \(X\sim\operatorname{Binomial}(10,0.60)\), then \(P(X=4)=\binom{10}{4}(0.60)^4(0.40)^6=0.111476736\).

Answer

a) There are \(15{,}504\) possible groups. The probability is \(\frac{385}{969}\approx39.73\%\). b) The probability is \(0.111476736\), or approximately \(11.15\%\).

52690012

A box contains \(15\) different chocolates, and \(5\) have nut filling. Four chocolates are selected at the same time. a) Find the probability that four of the five nut-filled chocolates are selected. b) Find the probability that no nut-filled chocolate is selected. Describe the sampling model.

Hints

- Separate the chocolates into nut-filled and not nut-filled categories. - Count all unordered four-chocolate samples. - For each event, decide how many chocolates must come from each category.

Solution

a) There are \(\binom{15}{4}=1365\) equally likely four-chocolate samples. For four nut-filled chocolates, \(P=\frac{\binom{5}{4}\binom{10}{0}}{\binom{15}{4}}=\frac{1}{273}\). b) For no nut-filled chocolates, \(P=\frac{\binom{5}{0}\binom{10}{4}}{\binom{15}{4}}=\frac{2}{13}\). Sampling model: select \(4\) distinct chocolates from \(15\) without replacement; order does not matter.

Answer

a) \(\frac{1}{273}\) b) \(\frac{2}{13}\) Sampling model: select \(4\) of \(15\) without replacement; order does not matter.

52692712

A shipment of \(40\) USB drives contains \(6\) defective drives. Five drives are selected at random without replacement for inspection. a) Find the probability that exactly two selected drives are defective. b) Find the probability that at most one selected drive is defective. c) Find the probability that at least one defective drive is found.

Hints

- Treat the five drives as an unordered sample without replacement. - For a fixed number of defectives, choose defective and nondefective drives separately. - Use case addition for “at most one” and a complement for “at least one.”

Solution

a) There are \(\binom{40}{5}\) equally likely samples. Thus, \(P(X=2)=\frac{\binom{6}{2}\binom{34}{3}}{\binom{40}{5}}=\frac{3740}{27417}\approx0.1364\). b) At most one defective drive means zero or one defective drive. Therefore, \(P(X\le1)=\frac{\binom{6}{0}\binom{34}{5}+\binom{6}{1}\binom{34}{4}}{\binom{40}{5}}=\frac{23188}{27417}\approx0.8458\). c) Use the complement of no defective drives: \(P(X\ge1)=1-\frac{\binom{34}{5}}{\binom{40}{5}}=\frac{15823}{27417}\approx0.5771\).

Answer

a) \(\frac{3740}{27417}\approx13.64\%\) b) \(\frac{23188}{27417}\approx84.58\%\) c) \(\frac{15823}{27417}\approx57.71\%\)

52692812

A class has \(24\) students, and \(10\) of them speak Spanish. Six students are selected at random for a group project. a) Find the probability that exactly three selected students speak Spanish. b) Find the probability that none of the selected students speaks Spanish. c) Find the probability that a majority of the group, more than three students, speaks Spanish.

Hints

- Treat the six students as an unordered sample without replacement. - For each case, choose Spanish speakers and non-Spanish speakers separately. - A majority in a group of six means \(4\), \(5\), or \(6\).

Solution

a) There are \(\binom{24}{6}\) equally likely six-student samples. Thus, \(P(X=3)=\frac{\binom{10}{3}\binom{14}{3}}{\binom{24}{6}}=\frac{1560}{4807}\approx0.3245\). b) For no Spanish speakers, \(P(X=0)=\frac{\binom{10}{0}\binom{14}{6}}{\binom{24}{6}}=\frac{39}{1748}\approx0.0223\). c) A majority means \(4\), \(5\), or \(6\) Spanish speakers. Therefore, \(P(X>3)=\frac{\binom{10}{4}\binom{14}{2}+\binom{10}{5}\binom{14}{1}+\binom{10}{6}\binom{14}{0}}{\binom{24}{6}}=\frac{816}{4807}\approx0.1698\).

Answer

a) \(\frac{1560}{4807}\approx32.45\%\) b) \(\frac{39}{1748}\approx2.23\%\) c) \(\frac{816}{4807}\approx16.98\%\)

52697612

An electronics store receives \(24\) smartphones, exactly \(3\) of which have a software defect. Four phones are selected at random for quality control. a) If the \(4\) phones are selected without replacement, find the probability that none has the defect. b) Find the probability that at least \(1\) selected phone has the defect. c) Find the probability from part a) if each phone were replaced before the next selection. Compare the results.

Hints

- 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.

Solution

a) Without replacement, \(P(X=0)=\frac{\binom{21}{4}}{\binom{24}{4}}=\frac{5985}{10626}\approx0.563241\). b) \(P(X\ge1)=1-P(X=0)\approx0.436759\). c) With replacement, the probability of a nondefective phone remains \(\frac{21}{24}=0.875\). Thus \(P(\text{four nondefective})=(0.875)^4\approx0.586182\), which is higher than without replacement.

Answer

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.

52705612

A club has \(12\) students, and \(7\) of them play a musical instrument. A random sample of \(4\) students is selected without replacement. The random variable \(X\) is the number of selected students who play an instrument. a) Create a probability distribution table for \(X\). b) Find the probability that more than half of the selected students play an instrument.

Hints

- Count the total number of samples of \(4\) students from \(12\). - For each value of \(X\), choose instrument players and nonplayers separately. - Translate “more than half” into a condition on \(X\). - Check that the probabilities in the table add to \(1\).

Solution

a) For \(k\in\{0,1,2,3,4\}\), \(P(X=k)=\frac{\binom{7}{k}\binom{5}{4-k}}{\binom{12}{4}}\), with \(\binom{12}{4}=495\). This gives probabilities \(\frac{1}{99},\frac{14}{99},\frac{42}{99},\frac{35}{99},\frac{7}{99}\) for \(k=0,1,2,3,4\), respectively. b) More than half of \(4\) means \(X>2\). Thus \(P(X>2)=\frac{35}{99}+\frac{7}{99}=\frac{14}{33}\approx0.4242\).

Answer

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) \(P(X>2)=\frac{14}{33}\approx42.42\%\).

52719912

A container holds \(6\) black balls and \(4\) white balls. Three balls are selected one at a time. A player wins by selecting at least \(2\) black balls. Determine whether sampling with replacement or without replacement gives the player the greater probability of winning.

Hints

- 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.

Solution

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.

Answer

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

52720012

A box contains \(50\) batteries, and \(5\) of them are dead. Four batteries are selected at random for a flashlight. a) Find the probability that all four selected batteries work if the selections are modeled as independent trials with replacement. b) Find the exact probability that all four selected batteries work when the batteries are actually selected without replacement. c) Compare the results and evaluate whether the binomial model is a reasonable approximation in this situation.

Hints

- Find the proportion of batteries that work. - Use independent repeated-trial reasoning for replacement and combinations for sampling without replacement. - Compare both the numerical results and the sample-size fraction when judging the approximation.

Solution

a) With replacement, the probability that a selected battery works is \(\frac{45}{50}=0.9\). Thus \(P(\text{all work})=(0.9)^4=0.6561\). b) Without replacement, \(P(\text{all work})=\frac{\binom{45}{4}}{\binom{50}{4}}=\frac{148995}{230300}\approx0.6470\). c) The binomial result is about \(0.0091\), or \(0.91\) percentage point, higher. The sample is \(\frac{4}{50}=8\%\) of the population, so it does not meet the \(5\%\) guideline used for approximating without-replacement sampling by independent trials. Numerically the two probabilities are fairly close, but the guideline itself does not justify the approximation here.

Answer

a) \(0.6561\) b) \(\frac{148995}{230300}\approx0.6470\) c) The binomial result is about \(0.0091\) higher. The sample is \(8\%\) of the population, so the \(5\%\) approximation guideline is not met, although the numerical difference is small.

52720412

A small choir has \(24\) members: \(14\) sopranos and \(10\) altos. Four members are selected at random to form a quartet. a) Find the probability that all four selected members are sopranos. b) Find the probability that the quartet contains exactly two sopranos and two altos. c) The choir has two featured soloists, both sopranos. Find the probability that both soloists are selected. d) Describe an urn model for this selection process and explain the parameters.

Hints

- Count all groups of \(4\) from \(24\). - For a group containing members from two voice parts, count the choices from each part separately. - If two specified people are already included, determine how many places remain to fill. - For the urn model, decide what each ball, color, special mark, and draw represents.

Solution

a) There are \(\binom{24}{4}=10{,}626\) possible quartets, so \(P(\text{all sopranos})=\frac{\binom{14}{4}}{\binom{24}{4}}=\frac{1001}{10626}\approx0.09420\). b) \(P(\text{two sopranos and two altos})=\frac{\binom{14}{2}\binom{10}{2}}{\binom{24}{4}}=\frac{4095}{10626}\approx0.38538\). c) If both specified soloists are selected, choose the other two members from the remaining \(22\): \(P=\frac{\binom{22}{2}}{\binom{24}{4}}=\frac{231}{10626}\approx0.02174\). d) Use \(24\) distinct balls, one for each choir member. Use one color for the \(14\) sopranos and another for the \(10\) altos, specially mark the two featured soloists, and draw \(4\) balls without replacement. Because the outcome is a quartet rather than an ordered sequence, ignore draw order.

Answer

a) \(\frac{1001}{10626}\approx0.09420\) b) \(\frac{4095}{10626}\approx0.38538\) c) \(\frac{231}{10626}\approx0.02174\) d) Use \(24\) distinct balls representing the members, encode voice part by color, mark the two soloists, and draw \(4\) without replacement while ignoring order.

52720912

A container holds \(15\) red balls, \(10\) green balls, and \(5\) blue balls. Four balls are selected at the same time without replacement. Find the probability of each event. \(E_1\): Exactly two red, one green, and one blue ball are selected. \(E_2\): No red ball is selected. \(E_3\): At least three blue balls are selected.

Hints

- Treat the four balls as an unordered sample without replacement. - Translate each event into the number selected from each color category. - For \(E_3\), identify the separate cases included in “at least three blue.”

Solution

\(E_1\): There are \(\binom{30}{4}\) equally likely four-ball samples. Thus, \(P(E_1)=\frac{\binom{15}{2}\binom{10}{1}\binom{5}{1}}{\binom{30}{4}}=\frac{50}{261}\approx0.1916\). \(E_2\): Choose all four balls from the \(15\) nonred balls: \(P(E_2)=\frac{\binom{15}{4}}{\binom{30}{4}}=\frac{13}{261}\approx0.0498\). \(E_3\): Add the cases of exactly three and exactly four blue balls: \(P(E_3)=\frac{\binom{5}{3}\binom{25}{1}+\binom{5}{4}\binom{25}{0}}{\binom{30}{4}}=\frac{17}{1827}\approx0.00930\).

Answer

\(E_1\): \(\frac{50}{261}\approx19.16\%\) \(E_2\): \(\frac{13}{261}\approx4.98\%\) \(E_3\): \(\frac{17}{1827}\approx0.93\%\)

53087712

An electronic lock uses a four-digit code, with each digit from \(0\) through \(9\). Two methods for randomly generating a new code are compared. Method A: A random-number generator selects each of the four digits independently, and every digit from \(0\) through \(9\) is equally likely at each position. Method B: An urn contains \(40\) balls, with each digit from \(0\) through \(9\) printed on exactly four balls. Four balls are drawn in order without replacement to form the code. 1. Find the probability that each method generates \(7777\). 2. Find the probability that each method generates \(1234\). 3. Use the ideas of independence and sampling with or without replacement to explain why Method B is not an exact model of Method A.

Hints

- 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.

Solution

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.

Answer

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.

53087812

A warehouse has \(500\) electronic components, including \(10\) defective components. A technician selects two components in sequence. 1. Find the probability that both components are defective if the first component is replaced before the second is selected. 2. Find the probability that both components are defective if the first component is not replaced. 3. For sampling without replacement, let \(E_1\) be “the first component is defective” and \(E_2\) be “the second component is defective.” Determine whether \(E_1\) and \(E_2\) are independent. 4. Decide whether sampling with replacement gives a useful approximation to sampling without replacement in this situation.

Hints

- 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.

Solution

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.

Answer

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.

53088412

A deck contains \(32\) cards, including exactly \(4\) aces. The deck is shuffled and the cards are revealed one at a time. a) What is the probability that the ace of hearts is exactly the fifth card revealed? b) Find the probability that exactly \(2\) of the first \(5\) cards are aces. c) Suppose each card were replaced and the deck reshuffled after every draw. Find the probability of drawing exactly \(2\) aces in \(5\) draws.

Hints

- 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.

Solution

a) Each card is equally likely to occupy the fifth position, so \(P(\text{ace of hearts fifth})=\frac{1}{32}=0.03125\). b) Without replacement, \(P(X=2)=\frac{\binom{4}{2}\binom{28}{3}}{\binom{32}{5}}=\frac{19656}{201376}\approx0.09761\). c) With replacement, each draw has ace probability \(p=\frac{4}{32}=\frac{1}{8}\). Therefore, \(P(X=2)=\binom{5}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^3\approx0.10468\).

Answer

a) \(\frac{1}{32}=0.03125\) b) \(\frac{\binom{4}{2}\binom{28}{3}}{\binom{32}{5}}\approx0.09761\) c) \(\binom{5}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^3\approx0.10468\)

53090212

A sports club has \(30\) members: \(12\) women and \(18\) men. A six-person team is selected at random. a) Find the probability that the team consists entirely of men. b) Find the probability that the team contains exactly \(3\) women and \(3\) men.

Hints

- Treat the team as an unordered sample without replacement. - Select the required numbers from the two groups separately. - Compare each favorable count with all six-person samples.

Solution

a) There are \(\binom{30}{6}\) equally likely six-person samples. For an all-men team, \(P=\frac{\binom{18}{6}}{\binom{30}{6}}\approx0.03126\). b) For exactly \(3\) women and \(3\) men, \(P=\frac{\binom{12}{3}\binom{18}{3}}{\binom{30}{6}}\approx0.30234\).

Answer

a) Approximately \(0.03126\), or \(3.13\%\) b) Approximately \(0.30234\), or \(30.23\%\)

53090912

A grade level has \(24\) students. Two different six-person work groups are selected independently. A student may belong to both groups. Find the probability that: a) exactly \(2\) students are in both groups; b) no student is in both groups.

Hints

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

Solution

a) Treat the first group of \(6\) as fixed. For the second group, choose \(2\) of those \(6\) and \(4\) of the other \(18\): \(P=\frac{\binom{6}{2}\binom{18}{4}}{\binom{24}{6}}\approx0.34102\). b) For no overlap, choose all \(6\) members of the second group from the other \(18\): \(P=\frac{\binom{18}{6}}{\binom{24}{6}}\approx0.13792\).

Answer

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

53094512

A box contains \(18\) chocolates: \(8\) with marzipan filling, \(6\) with nougat filling, and \(4\) with caramel filling. Three chocolates are selected without replacement. Find the probability that: 1. All three chocolates have the same filling. 2. The three chocolates have three different fillings.

Hints

- Treat the three chocolates as an unordered sample without replacement. - For the same filling, count each filling type separately and add. - For three different fillings, choose one chocolate from each category.

Solution

There are \(\binom{18}{3}=816\) equally likely three-chocolate samples. 1. For the same filling, add the samples of three marzipan, three nougat, or three caramel chocolates: \(P=\frac{\binom{8}{3}+\binom{6}{3}+\binom{4}{3}}{\binom{18}{3}}=\frac{5}{51}\). 2. For three different fillings, choose one chocolate of each type: \(P=\frac{\binom{8}{1}\binom{6}{1}\binom{4}{1}}{\binom{18}{3}}=\frac{4}{17}\).

Answer

1. \(\frac{5}{51}\approx9.8\%\) 2. \(\frac{4}{17}\approx23.5\%\)

53094712

In a school raffle game, four numbers are drawn without replacement from \(25\) numbered balls. A player marks four numbers before the drawing. Find the probability that the player matches: 1. Exactly four numbers. 2. Exactly two numbers. 3. At least three numbers.

Hints

- Separate the \(25\) numbers into the player's four choices and the other \(21\) numbers. - For an exact number of matches, choose from both groups. - “At least three” combines the cases of three and four matches.

Solution

1. There are \(\binom{25}{4}=12{,}650\) possible four-number sets. For exactly four matches, \(P(X=4)=\frac{\binom{4}{4}\binom{21}{0}}{\binom{25}{4}}=\frac{1}{12650}\). 2. For exactly two matches, \(P(X=2)=\frac{\binom{4}{2}\binom{21}{2}}{\binom{25}{4}}=\frac{126}{1265}\approx 0.0996\). 3. At least three matches means exactly three or exactly four, so \(P(X\ge 3)=\frac{\binom{4}{3}\binom{21}{1}+\binom{4}{4}\binom{21}{0}}{\binom{25}{4}}=\frac{17}{2530}\approx 0.00672\).

Answer

1. \(\frac{1}{12650}\approx 0.0079\%\) 2. \(\frac{126}{1265}\approx 9.96\%\) 3. \(\frac{17}{2530}\approx 0.672\%\)

53095412

An urn contains \(10\) white balls and \(10\) black balls. Three balls are selected in sequence. 1. Find the probability of selecting exactly \(2\) white balls without replacement. 2. Find the probability of selecting exactly \(2\) white balls with replacement. 3. Compare the probabilities and explain the difference between the without-replacement and with-replacement models.

Hints

- 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.

Solution

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 \(0.5\).

Answer

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.

53095512

A fair \(12\)-sided die labeled \(1\) through \(12\) is rolled three times. 1. Find the probability that all three results are different. 2. Find the probability that the results form a strictly increasing sequence, such as \((2, 5, 11)\). 3. The random variable \(X\) is the number of distinct values among the three rolls. Create a probability distribution table for \(X\).

Hints

- 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 \(1\).

Solution

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 \(P(X=3)=\frac{55}{72}\).

Answer

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>

53117112

In a city with \(25{,}000\) households, \(15\%\) subscribe to a particular local newspaper. A market-research study randomly selects \(50\) households without replacement. a) Explain why the probability of selecting a subscriber can still be treated as approximately constant from one selection to the next. b) Use an appropriate probability distribution to approximate the probability that exactly \(8\) of the \(50\) selected households subscribe.

Hints

- 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.

Solution

a) 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. b) Use \(X\sim\operatorname{Binomial}(50,0.15)\). Then \(P(X=8)=\binom{50}{8}(0.15)^8(0.85)^{42}\approx0.14935\).

Answer

a) The sample is only \(0.2\%\) of the population, so sampling without replacement changes the probability by a negligible amount. b) \(P(X=8)\approx0.14935\), or about \(14.93\%\).

53117212

An electronics store receives a shipment of \(1200\) memory chips. Historically, \(3\%\) of the chips are defective. An employee inspects a sample selected without replacement. a) Using the condition \(\frac{n}{N}\le0.05\), find the largest sample size \(n\) for which a binomial distribution may reasonably approximate the sampling process. b) For a sample of \(30\) chips, use the binomial approximation to find the probability that at most one chip is defective.

Hints

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

Solution

a) The condition requires \(n\le0.05(1200)=60\), so the largest allowed sample size is \(60\). b) Let \(X\sim\operatorname{Binomial}(30,0.03)\). Then \(P(X\le1)=P(X=0)+P(X=1)=(0.97)^{30}+\binom{30}{1}(0.03)(0.97)^{29}\approx0.77308\).

Answer

a) \(n=60\) b) \(P(X\le1)\approx0.77308\), or about \(77.31\%\).

53118912

At a nursery, \(85\%\) of the seedlings of a certain variety typically bloom for more than three months. a) A landscape designer buys \(12\) seedlings for a park. Assuming a binomial model, find the probability that more than \(10\) of them bloom for more than three months. b) A display contains only \(20\) seedlings of this variety, and exactly \(17\) are known to have the longer blooming period. A customer randomly selects \(5\) seedlings without replacement. Find the probability that all \(5\) selected seedlings have the longer blooming period. c) Explain the essential difference between the models in parts a) and b).

Hints

- 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.

Solution

a) Let \(X\sim\operatorname{Binomial}(12,0.85)\). Then \(P(X>10)=P(X=11)+P(X=12)=\binom{12}{11}(0.85)^{11}(0.15)+(0.85)^{12}\approx0.44346\). b) Sampling is without replacement from a finite set, so \(P(X=5)=\frac{\binom{17}{5}}{\binom{20}{5}}=\frac{6188}{15504}\approx0.39912\). c) Part a) treats each seedling as an independent trial with constant success probability \(0.85\). In part b), the sample is a substantial fraction of a small fixed group, so each selection changes later probabilities.

Answer

a) \(P(X>10)\approx0.44346\), or about \(44.35\%\) b) \(\frac{6188}{15504}\approx0.39912\), 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.

53119112

An urn contains \(20\) balls labeled \(1\) through \(20\). Three balls are drawn without replacement. The random variable \(X\) is the number of selected balls labeled from \(1\) through \(5\). Determine the probability distribution of \(X\) in a table and calculate \(E(X)\).

Hints

- Identify the possible values of \(X\) and count selections with exactly \(k\) target balls. - After building the distribution, multiply each possible value of \(X\) by its probability. - Add those weighted values to obtain the expected value.

Solution

1. There are \(5\) target balls and \(15\) other balls. For \(k\in\{0,1,2,3\}\), \(P(X=k)=\frac{\binom{5}{k}\binom{15}{3-k}}{\binom{20}{3}}\), with \(\binom{20}{3}=1140\). 2. This gives \(P(X=0)=\frac{91}{228}\), \(P(X=1)=\frac{35}{76}\), \(P(X=2)=\frac{5}{38}\), and \(P(X=3)=\frac{1}{114}\). 3. Use the distribution to compute \(E(X)=0\cdot\frac{91}{228}+1\cdot\frac{35}{76}+2\cdot\frac{5}{38}+3\cdot\frac{1}{114}=\frac{3}{4}=0.75\). 4. As a check, the expected count also equals the sample size times the target proportion: \(3\cdot\frac{5}{20}=0.75\).

Answer

<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> \(E(X)=0.75\)

53202612

Use the urn shown. Two balls are drawn. a) Find the probability that the balls have the same color when the first ball is replaced before the second draw. b) Find the probability that the balls have the same color when the first ball is not replaced. c) In which case is the probability of drawing at least one red ball greater? Justify your answer with calculations.

Figure for problem 532026

Hints

- Read the red and blue counts from the urn. - Write the two same-color paths for each sampling method. - For part c, use the complement of drawing two blue balls.

Solution

a) With replacement, \(P(\text{same color})=\left(\frac{4}{10}\right)^2+\left(\frac{6}{10}\right)^2=\frac{13}{25}=0.52\). b) Without replacement, \(P(\text{same color})=\frac{4}{10}\cdot\frac{3}{9}+\frac{6}{10}\cdot\frac{5}{9}=\frac{7}{15}\). c) With replacement, \(P(\text{at least one red})=1-\left(\frac{6}{10}\right)^2=\frac{16}{25}\). 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.

Answer

a) \(\frac{13}{25}=52\%\) b) \(\frac{7}{15}\approx46.7\%\) c) Without replacement: \(\frac{2}{3}\approx66.7\%\), compared with \(\frac{16}{25}=64\%\) with replacement.

53211712

Use the urn shown. Three balls are drawn in sequence. a) Find the probability of drawing the sequence “NAS” with replacement. b) Find the probability of drawing the sequence “NAS” without replacement. c) Three balls are drawn without replacement. Find the probability that the drawn letters can be rearranged to spell “NAS.”

Figure for problem 532117

Hints

- Read how many balls show each letter in the urn. - Decide whether the letter probabilities remain constant after each draw. - For part c, count the different orders of one N, one A, and one S.

Solution

a) With replacement, \(P(\text{NAS})=\frac{2}{6}\cdot\frac{3}{6}\cdot\frac{1}{6}=\frac{1}{36}\). b) Without replacement, \(P(\text{NAS})=\frac{2}{6}\cdot\frac{3}{5}\cdot\frac{1}{4}=\frac{1}{20}\). c) 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 \(\frac{1}{20}\), so the probability is \(6\cdot\frac{1}{20}=\frac{3}{10}\).

Answer

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

53213812

Use the urn shown. Two balls are drawn without replacement. A player earns \(\$2\) for each B ball, earns \(\$1\) for each R ball, and loses \(\$3\) for each Y ball. The random variable \(X\) is the player's total gain or loss after two draws. Determine the probability distribution of \(X\) and present it in a table.

Figure for problem 532138

Hints

- Read the R, B, and Y counts from the urn. - List every possible pair of labels and its total gain or loss. - Add probabilities of different draw orders that produce the same value of \(X\). - Check that the distribution probabilities add to \(1\).

Solution

1. There are \(9\cdot8=72\) equally likely ordered draws when the physical balls are distinguished. 2. The ordered-path counts are \(2\) for \(X=-6\), \(12\) for \(X=-2\), \(16\) for \(X=-1\), \(6\) for \(X=2\), \(24\) for \(X=3\), and \(12\) for \(X=4\). 3. Dividing each count by \(72\) gives the probability distribution.

Answer

<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>

53612412

Use the urn shown. Two balls are drawn. Determine whether the probability of drawing two balls of different colors is greater with replacement or without replacement. Support your conclusion with calculations.

Figure for problem 536124

Hints

- Read the two color counts from the urn. - List the two orders that produce different colors. - Calculate the two sampling methods separately.

Solution

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

Answer

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

53615312

Use the urn shown. Two balls are drawn in order without replacement. Find the probability that the second number is greater than the first number.

Figure for problem 536153

Hints

- Read the available labels from the urn. - Consider the two possible orders for each pair of different numbers. - Equality cannot occur.

Solution

1. There are \(5\cdot4=20\) equally likely ordered pairs of different numbers. 2. For every unordered pair, exactly one order has the second number greater than the first, and the other has the first number greater than the second. 3. Equality is impossible, so the two cases are equally likely. Therefore, \(P(\text{second greater than first})=\frac{1}{2}\).

Answer

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

53618112

Use the urn shown. a) Three balls are drawn with replacement. Find the probability of drawing the sequence “SIS.” b) Three balls are drawn without replacement. Find the probability of drawing the sequence “TIT.” c) Three balls are drawn without replacement. Find the probability of drawing A, K, and T in any order.

Figure for problem 536181

Hints

- Read the multiplicity of each letter from the urn. - Check whether each part uses replacement. - For a specified sequence, multiply probabilities in order. - For part c, count the possible orders of A, K, and T.

Solution

a) With replacement, \(P(\text{SIS})=\frac{2}{9}\cdot\frac{2}{9}\cdot\frac{2}{9}=\frac{8}{729}\). b) Without replacement, \(P(\text{TIT})=\frac{3}{9}\cdot\frac{2}{8}\cdot\frac{2}{7}=\frac{1}{42}\). c) The letters A, K, and T can occur in \(3!=6\) orders. Each order has probability \(\frac{1}{9}\cdot\frac{1}{8}\cdot\frac{3}{7}=\frac{1}{168}\), so the total probability is \(6\cdot\frac{1}{168}=\frac{1}{28}\).

Answer

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

53754712

Use diagrams a) and b) to compare the probability of drawing the gold-colored ball at least once in two draws. Diagram a) shows drawing with replacement, and diagram b) shows drawing without replacement.

Figure for problem 537547

Hints

- In each diagram, use the complement of drawing only gray balls. - Compare the second gray branch after a gray first draw.

Solution

1. In diagram a), use the complement of two gray draws: \(1-\left(\frac{4}{5}\right)^2=\frac{9}{25}=0.36\). 2. In diagram b), the complement probability is \(1-\frac{4}{5}\cdot\frac{3}{4}=\frac{2}{5}=0.40\). 3. The without-replacement probability is greater by \(0.40-0.36=0.04\).

Answer

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

52691412

A parts manufacturer ships boxes of \(150\) components. One box contains exactly \(30\) defective components. A quality-control sample of \(10\) components is selected. a) Find the probability that exactly \(2\) selected components are defective if each component is replaced before the next selection. b) Find the probability that at least \(1\) selected component is defective if sampling is with replacement. c) The actual sample is selected without replacement. Explain whether the probability of selecting no defective components is greater or less without replacement than with replacement.

Hints

- 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 chance of another nondefective component changes after one is removed.

Solution

a) With replacement, \(p=\frac{30}{150}=0.20\). Thus \(P(X=2)=\binom{10}{2}(0.20)^2(0.80)^8\approx0.301990\). b) \(P(X\ge1)=1-(0.80)^{10}\approx0.892626\). c) Without replacement, after a nondefective component is selected, the next conditional probability of another nondefective component is slightly less than \(0.80\). Therefore, the probability of selecting no defective components is less without replacement than with replacement.

Answer

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.

52697512

A container holds \(15\) tickets: \(5\) winning tickets and \(10\) losing tickets. a) Six tickets are drawn with replacement. Find the probability that exactly \(2\) winning tickets are drawn. b) Six tickets are selected at once without replacement. Find the probability that exactly \(2\) are winning tickets. c) What is the minimum number of draws with replacement needed so that the probability of drawing at least \(1\) winning ticket is greater than \(95\%\)?

Hints

- 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. - Account for the sign of the logarithm when solving the inequality.

Solution

a) Let \(X\sim\operatorname{Binomial}(6,\frac{1}{3})\). Then \(P(X=2)=\binom{6}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^4=\frac{80}{243}\approx0.329218\). b) Without replacement, \(P(X=2)=\frac{\binom{5}{2}\binom{10}{4}}{\binom{15}{6}}=\frac{420}{1001}\approx0.419580\). c) Require \(1-\left(\frac{2}{3}\right)^n>0.95\), so \(\left(\frac{2}{3}\right)^n<0.05\). This gives \(n>\frac{\ln(0.05)}{\ln(2/3)}\approx7.388\), so the minimum is \(8\).

Answer

a) \(\frac{80}{243}\approx32.92\%\) b) \(\frac{420}{1001}\approx41.96\%\) c) At least \(8\) draws are needed.

52705812

An electronics retailer receives two battery shipments. Shipment 1 contains \(50\) batteries, \(5\) of which are defective. Shipment 2 contains \(500\) batteries, \(50\) of which are defective. A random sample of \(4\) batteries is selected from one shipment. a) Approximate the probability that exactly \(1\) sampled battery is defective by modeling the selections as Bernoulli trials with replacement. b) Find the exact probability of exactly \(1\) defective battery for each shipment when sampling without replacement. c) A common guideline says the binomial distribution is a good approximation to sampling without replacement when the sample size is at most \(5\%\) of the population. Check the guideline for each shipment and evaluate the approximation using your answers.

Hints

- 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.

Solution

a) Both shipments have defect proportion \(p=0.10\). The binomial approximation is \(\binom{4}{1}(0.10)(0.90)^3=0.2916\). b) For Shipment 1, the exact probability is \(\frac{\binom{5}{1}\binom{45}{3}}{\binom{50}{4}}\approx0.308076\). For Shipment 2, it is \(\frac{\binom{50}{1}\binom{450}{3}}{\binom{500}{4}}\approx0.293164\). c) For Shipment 1, \(4>0.05(50)=2.5\), so the guideline is not met and the approximation differs noticeably from the exact value. For Shipment 2, \(4\le0.05(500)=25\), so the guideline is met and the approximation is close to the exact value.

Answer

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.

53088012

A container holds \(10\) balls numbered \(1\) through \(10\). Three balls are selected without replacement. a) Find the probability that ball \(1\) is selected. b) Find the probability that at least one of balls \(1\) and \(2\) is selected. c) Tariq claims, “The probability of selecting one particular ball is \(3\cdot\frac{1}{10}\), so the probability of selecting at least one of two particular balls must be \(3\cdot\frac{2}{10}=0.6\).” Use your results to evaluate the claim and explain the error.

Hints

- Count unordered selections of \(3\) balls 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?

Solution

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, so the probability is \(\frac{36}{120}=\frac{3}{10}=0.3\). 2. For the complement in part b), select all three balls from the other eight: \(\binom{8}{3}=56\) selections. Thus the probability of selecting at least one of balls \(1\) and \(2\) is \(1-\frac{56}{120}=\frac{8}{15}\approx0.533\). 3. Tariq’s claim is false because \(\frac{8}{15}\ne0.6\). The events “ball \(1\) is selected” and “ball \(2\) is selected” overlap because both balls can appear in one sample. Adding the two probabilities without subtracting the overlap double-counts those samples.

Answer

a) \(0.3\) b) \(\frac{8}{15}\approx0.533\) c) The claim is false. The two events are not disjoint, so adding their probabilities double-counts selections containing both balls.

53753012

A bag contains eight balls, all red or blue. Two balls are drawn without replacement. The probability of drawing red first and blue second is \(\frac{15}{56}\). Find all possible compositions of the bag.

Hints

- Let a variable represent the number of red balls and express the number of blue balls in terms of it. - Write the probability of drawing red first and blue second from the unknown composition. - Solve the resulting equation and keep only feasible integer ball counts.

Solution

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

Answer

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

53753112

A bag contains ten balls, including \(x\) red balls. Two balls are drawn without replacement. The probability that both balls are red is \(\frac{2}{15}\). Find \(x\).

Hints

- Express the probability of two red draws in terms of the unknown red count. - After one red ball is drawn, both the red count and total count change. - Solve the resulting equation and reject values that cannot represent a number of balls.

Solution

1. The first red probability is \(\frac{x}{10}\). After one red ball is drawn, \(x-1\) red balls remain among \(9\) balls. 2. Therefore, \(\frac{x}{10}\cdot\frac{x-1}{9}=\frac{2}{15}\). 3. Simplifying gives \(x(x-1)=12\), so \(x^2-x-12=0\). 4. Factoring gives \((x-4)(x+3)=0\). The only feasible ball count is \(x=4\).

Answer

\(x=4\)

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