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Sampling with and without replacement

Learning goals

Distinguishes sampling with replacement from sampling without replacement.
Determines whether successive selections are independent.
Updates probabilities after selections without replacement.
Builds tree diagrams or products that match the sampling process.

Difficulty: 1 – 5

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Sampling with and without replacement 1

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

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

Sampling with and without replacement 2

A bag contains 4 green balls and 1 yellow ball. One ball is drawn at random, its color is recorded, and the ball is returned. The bag is mixed again before a second random draw, so each ball is equally likely to be selected. What is the probability that the second ball is yellow?

Hints

- Focus on what the bag contains immediately before the second draw. - Does knowing the color of the first draw change the chance on the second draw here? Why or why not?

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

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

Sampling with and without replacement 3

One B ball is removed from the urn. A second ball is drawn, with each remaining ball equally likely. What is the probability that the second ball 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

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

Sampling with and without replacement 4

A bag contains equal numbers of green and orange balls. A green ball is drawn first and is not replaced. Each ball in the bag is equally likely to be selected at each draw. 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.

Sampling with and without replacement 5

A jar has red and white marbles. A marble is drawn at random, its color is recorded, and it is returned to the jar. The jar is mixed before a second random draw. Is the color of the second draw independent of the color of the first? What would change if the first marble were kept out?

Hints

- Is the jar's makeup the same before each draw? - If one marble is removed, what information about it affects the jar for the second draw?

Solution

1. Returning the marble and mixing restores the same color proportions, so the probability on the second draw does not depend on the first color. 2. The draw colors are independent with replacement. 3. If the first marble were kept out, the remaining color proportions would depend on what was removed, so the second-draw probability would generally change.

Answer

Yes, with replacement the draw colors are independent. Without replacement, knowing the first color generally changes the chance of the second color.

Sampling with and without replacement 6

A bag has two red beads and four blue beads. One red bead has been drawn and kept out. A second bead is drawn at random, with each remaining bead equally likely. What is the probability that the second bead is red?

Hints

- Which color count was reduced by the first draw? - How many beads remain altogether before the second draw?

Solution

1. After the red bead is removed, one red bead and four blue beads remain. 2. There are five beads left, so the probability of red on the second draw is \(\frac{1}{5}\).

Answer

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

Sampling with and without replacement 7

A bag starts with two red balls and three yellow balls. One red ball is drawn at random and kept out. A student says, “The probability that the next ball is yellow is still \(\frac{3}{5}\).” What is wrong with the student's fraction? Give the correct probability if each remaining ball is equally likely.

Hints

- Which count stayed unchanged after the first draw? - How many balls are actually available for the next draw?

Solution

1. The yellow count is still \(3\), but only \(4\) balls remain after one red ball is removed. 2. The correct second-draw probability is \(\frac{3}{4}\), not \(\frac{3}{5}\).

Answer

The denominator must change. The correct probability is \(\frac{3}{4}\).

Sampling with and without replacement 8

A box contains \(20\) tickets, including \(5\) winning tickets. Four tickets are drawn in sequence. Each draw chooses uniformly from the tickets currently in the box. 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.

Sampling with and without replacement 9

A box contains \(24\) chocolates. Four have marzipan filling, and the rest have nougat filling. Two chocolates are drawn without replacement, with each remaining chocolate equally likely to be drawn next. Find the probability that both chocolates have nougat filling.

Hints

- How many chocolates have the filling you want before either draw? - If the first chocolate has nougat filling, what is left in the box? - How can you combine the chance of the first draw with the chance of the second, given that the first was nougat?

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

\(\frac{95}{138}\approx0.6884\)

Sampling with and without replacement 10

A box contains \(12\) chocolates: \(8\) are coated in dark chocolate and \(4\) are coated in milk chocolate. Three chocolates are drawn without replacement, with each remaining chocolate equally likely to be drawn next. 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

- What is different about the possible draws after you have removed one chocolate? - How many chocolates of the required coating would remain after each successful draw? - How does the chance of the next draw depend on the earlier ones?

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) \(\frac{1}{55}\)

Sampling with and without replacement 11

A bag contains three gold-colored coins and five silver-colored coins. Two coins are drawn without replacement, with each remaining coin equally likely to be drawn next. 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

- If the first coin is gold-colored, what is left for the second draw? - What different first-draw results could still lead to a matching-color pair? - Could one draw order be counted twice?

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) \(\frac{13}{28}\)

Sampling with and without replacement 12

Two balls are drawn from the urn without replacement. Each remaining ball is equally likely to be selected next. 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) \(\frac{2}{15}\)

Sampling with and without replacement 13

Two balls are drawn from the urn without replacement, with each remaining ball equally likely to be selected next. Find the probability that the second ball is green. Show a calculation that includes both possible colors of the first ball.

Figure for problem 536115

Hints

- How does the chance of green on the second draw depend on the first color? - What can happen on the first draw if the second draw is green? - How could you check that you have included all the ways for the second draw to be green?

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

\(\frac{3}{5}=0.6\). The two cases are green then green, with probability \(\frac{3}{10}\), and orange then green, also with probability \(\frac{3}{10}\). Their sum is \(\frac{3}{5}\).

Sampling with and without replacement 14

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

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

Sampling with and without replacement 15

Two probability trees represent two draws from a bag. Match each diagram with “with replacement” or “without replacement,” and explain 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; the second-draw probabilities stay at \(\frac{3}{5}\) and \(\frac{2}{5}\) after either first draw. B: without replacement; the second-draw probabilities change depending on the first ball, and only four balls remain.

Sampling with and without replacement 16

A partial probability tree shows the branches needed to study two card draws without replacement. Find the probability that both cards are aces. Give your answer as a fraction.

Figure for problem 537532

Hints

- Which first-draw result makes two aces possible? - What does the second branch's probability describe, given the first card?

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

\(\frac{1}{28}\)

Sampling with and without replacement 17

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

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

Sampling with and without replacement 18

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

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

Sampling with and without replacement 19

The probability tree describes two token draws without replacement. Every remaining token is equally likely to be drawn. Give the missing probability for each of the four second-draw branches. Write your answers as first outcome–second outcome followed by the probability.

Figure for problem 561663

Hints

- Read the starting proportions and total from the first stage. - After one outcome, which kind of token was removed? - The branches leaving the same node should add to one.

Solution

1. The first stage shows \(2\) star tokens and \(3\) circle tokens out of \(5\). 2. After a star, one star and three circles remain among four tokens, giving \(P(S\mid S)=\frac{1}{4}\) and \(P(C\mid S)=\frac{3}{4}\). 3. After a circle, two stars and two circles remain among four tokens, giving \(P(S\mid C)=\frac{2}{4}\) and \(P(C\mid C)=\frac{2}{4}\).

Answer

Star–star: \(\frac{1}{4}\); star–circle: \(\frac{3}{4}\); circle–star: \(\frac{1}{2}\); circle–circle: \(\frac{1}{2}\).

Sampling with and without replacement 20

Eight cards numbered \(1\) through \(8\) are shuffled. Two cards are drawn at random without replacement, with all remaining cards equally likely. A student finds the chance that both numbers are even by calculating \(\left(\frac{4}{8}\right)^2\). Explain why this method is incorrect, and find the correct probability.

Hints

- What must have happened on the first draw if both cards are even? - How many even cards and total cards remain after that result? - Should the probability for the second draw be the same as for the first?

Solution

1. Four of the eight cards have even numbers, so the first even-draw probability is \(\frac{4}{8}\). 2. If the first card is even, only three even cards remain among seven cards. The second conditional probability is \(\frac{3}{7}\), not \(\frac{4}{8}\). 3. The chance of two even numbers is \(\frac{4}{8}\cdot\frac{3}{7}=\frac{3}{14}\).

Answer

\(\frac{3}{14}\). The student repeats the original probability even though the first even card was not replaced.

Sampling with and without replacement 21

A box holds two gold tokens and four plain tokens. Two tokens are chosen at random. Every available token is equally likely at each draw. The chance of gold on the first draw is the same whether the first token is replaced or not. Compare the probability of drawing two gold tokens under the two rules. Why do the final probabilities differ?

Hints

- Does the first draw depend on a previous draw? - After one gold token appears, what remains in each version of the box? - Compare the second factors of the two probability products.

Solution

1. In both models, the first-draw chance of gold is \(\frac{2}{6}=\frac{1}{3}\). 2. With replacement, the next gold chance is again \(\frac{2}{6}\), so \(P(GG)=\frac{2}{6}\cdot\frac{2}{6}=\frac{1}{9}\). 3. Without replacement, one gold has been removed, so \(P(GG)=\frac{2}{6}\cdot\frac{1}{5}=\frac{1}{15}\). 4. The first-draw chances agree, but the second-draw chances differ because removing a token changes the box.

Answer

With replacement: \(\frac{1}{9}\). Without replacement: \(\frac{1}{15}\). The second draw is affected by the first only without replacement.

Sampling with and without replacement 22

A common rule of thumb says that sampling without replacement can sometimes be treated as approximately independent when the sample is no more than \(5\%\) of the population. A student samples \(15\) names from a school of \(600\) students. Another samples \(15\) names from a club of \(150\) students. a) Which sample meets this guideline? Show the sample fractions. b) Compare what can happen to a name already drawn under the two methods: with replacement and without replacement. Explain why meeting the guideline cannot make the two models identical.

Hints

- Divide the sample size by its population size in each case. - Compare the two fractions with the stated cutoff. - What happens to a student's name once it has been drawn without replacement? - Can changing the population size make an impossible repeated draw possible?

Solution

1. For the school, the fraction is \(\frac{15}{600}=0.025=2.5\%\). 2. For the club, the fraction is \(\frac{15}{150}=0.10=10\%\). 3. The school sample meets the \(5\%\) guideline; the club sample does not. 4. Without replacement, drawing the same student twice has probability \(0\). With replacement it has a positive probability. The guideline may support an approximation for some events but cannot make the two sampling rules identical.

Answer

a) School: \(2.5\%\), meets the guideline. Club: \(10\%\), does not. b) No. A repeated name is impossible without replacement but possible with replacement, so the models are not identical.

Sampling with and without replacement 23

A work group has \(6\) women and \(4\) men. Three people are drawn one at a time without replacement for a presentation. Each person remaining is equally likely to be drawn next. 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) \(\frac{29}{30}\approx0.967\)

Sampling with and without replacement 24

A standard \(52\)-card deck contains exactly four aces. Three cards are drawn without replacement, with every remaining card equally likely to be drawn next. 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, \(P(\text{exactly one ace})=3\cdot\frac{376}{5525}=\frac{1128}{5525}\approx0.2042\).

Answer

\(\frac{1128}{5525}\approx0.2042\)

Sampling with and without replacement 25

A crate contains \(20\) apples, and \(4\) of them are bruised. Three apples are selected without replacement so that every possible group of three apples is equally likely. 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) \(\frac{8}{95}\)

Sampling with and without replacement 26

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.

Sampling with and without replacement 27

A bag contains four yellow balls and six blue balls. Two balls are drawn without replacement, with each remaining ball equally likely to be drawn next. a) Find the probability that the two balls have different colors. b) Find the probability that at least one ball is yellow.

Hints

- Does getting two different colors require a particular color on the first draw? - Remember that the total number of balls changes after the first draw. - Is it easier to count outcomes with a yellow ball or outcomes with no yellow ball?

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) \(\frac{2}{3}\)

Sampling with and without replacement 28

A class has \(20\) students. Twelve own a tablet and \(8\) do not. a) Five students are selected without replacement so that every five-person group is equally likely. 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, a total of \(10\) independent draws are made. On each draw, one of the \(20\) students is equally likely to be selected, and the student's name is returned before the next draw. Find the probability that exactly \(4\) of the \(10\) draws select a tablet owner.

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 \(\frac{6160}{15504}=\frac{385}{969}\approx0.397317\). 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 \(11.15\%\).

Sampling with and without replacement 29

A box contains \(15\) different chocolates, and \(5\) have nut filling. Four chocolates are selected at the same time so that every possible group of four chocolates is equally likely. 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.

Sampling with and without replacement 30

A shipment of \(40\) USB drives contains \(6\) defective drives. Five drives are selected without replacement for inspection, with every possible group of five drives equally likely. 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

- Are two samples different if they contain the same drives in different orders? - Which exact numbers of defective drives satisfy “at most one”? - Which result is easier to count: finding at least one defective drive or finding none?

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) \(\frac{15823}{27417}\approx57.71\%\)

Sampling with and without replacement 31

A class has \(24\) students, and \(10\) of them speak Spanish. Six students are selected for a group project so that every six-person group is equally likely. 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) \(\frac{816}{4807}\approx16.98\%\)

Sampling with and without replacement 32

An electronics store receives \(24\) smartphones, exactly \(3\) of which have a software defect. Four phones are selected for quality control. At each draw, every phone currently available is equally likely to be selected. 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.

Sampling with and without replacement 33

A club has \(12\) students, and \(7\) of them play a musical instrument. Four students are selected without replacement so that every possible group is equally likely. 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

- What are the smallest and largest possible values of \(X\)? - Can you describe a four-person group for each value of \(X\)? - Which values would mean that a majority of the group plays an instrument? - How can you check that no possible group has been missed?

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

Sampling with and without replacement 34

A container holds \(6\) black balls and \(4\) white balls. Three balls are drawn one at a time; each available ball is equally likely to be selected at each draw. 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

- What happens to the chance of black on later draws under each rule? - Which outcomes count as a win? - Can you describe a sample space suited to each of the two rules? - Compare the two resulting chances, not just the number of black balls available at the start.

Solution

1. With replacement, let \(X\sim\operatorname{Binomial}(3,0.60)\). 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 \(\frac{80}{120}=\frac{2}{3}\approx0.666667\). 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.

Sampling with and without replacement 35

A box contains \(50\) batteries, and \(5\) of them are dead. Four batteries are drawn for a flashlight, with each available battery equally likely to be selected at each draw. 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. A common rule of thumb is that a binomial model often gives a good approximation to sampling without replacement when the sample is at most \(5\%\) of the population. This is a guideline, not a guarantee. Does the sample meet the guideline, and how close are the two probabilities?

Hints

- Find the proportion of batteries that work. - How does replacing a battery, or not replacing it, affect the next selection? - 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.

Sampling with and without replacement 36

A small choir has \(24\) members: \(14\) sopranos and \(10\) altos. Four members are selected to form a quartet so that every possible four-person group is equally likely. 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.

Sampling with and without replacement 37

A container holds \(15\) red balls, \(10\) green balls, and \(5\) blue balls. Four balls are selected at the same time without replacement so that every possible four-ball group is equally likely. 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

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

Sampling with and without replacement 38

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. At each draw, every remaining ball is equally likely to be chosen. 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, \(P_B(7777)=\frac{4}{40}\cdot\frac{3}{39}\cdot\frac{2}{38}\cdot\frac{1}{37}\approx0.00001094\). 2. In Method A, \(P_A(1234)=\left(\frac{1}{10}\right)^4=0.0001\). In Method B, \(P_B(1234)=\frac{4}{40}\cdot\frac{4}{39}\cdot\frac{4}{38}\cdot\frac{4}{37}\approx0.00011672\). 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.

Sampling with and without replacement 39

A box holds \(500\) batteries, including \(10\) faulty batteries. Two batteries are drawn one after the other. Every available battery is equally likely to be picked. 1. Find the probability that both batteries are faulty if the first one is replaced before the second draw. 2. Find the probability that both batteries are faulty if the first one is not replaced. 3. Without replacement, let \(E_1\) mean “the first battery is faulty” and \(E_2\) mean “the second battery is faulty.” Are \(E_1\) and \(E_2\) independent? 4. Compare the absolute difference and the relative error of the with-replacement result, using the without-replacement probability as the exact value. Is the approximation accurate for this event?

Hints

- Compare the second-draw probability with and without replacement. - Independence requires \(P(A\cap B)=P(A)P(B)\). - Compare the probabilities first by subtraction, then by expressing the difference as a fraction of the exact probability.

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, \(P(\text{both defective})=\frac{10}{500}\cdot\frac{9}{499}=\frac{90}{249500}\approx0.0003607\). 3. By symmetry, \(P(E_1)=P(E_2)=0.02\). However, \(P(E_1\cap E_2)\approx0.0003607\), while \(P(E_1)P(E_2)=0.0004\). Because these values are unequal, the events are dependent. 4. The absolute difference is \(0.0004-\frac{90}{249500}\approx0.0000393\). Relative to the exact probability, the error is about \(10.9\%\). The approximation is close on an absolute scale but noticeably inaccurate as a percentage of this rare event's probability. Whether it is useful depends on the required accuracy.

Answer

1. \(0.0004\) 2. \(\frac{90}{249500}\approx0.0003607\) 3. The events are dependent because \(P(E_1\cap E_2)\ne P(E_1)P(E_2)\). 4. Absolute difference approximately \(0.0000393\); relative error about \(10.9\%\). The approximation is not very accurate in relative terms.

Sampling with and without replacement 40

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) \(\binom{5}{2}\left(\frac{1}{8}\right)^2\left(\frac{7}{8}\right)^3\approx0.10468\)

Sampling with and without replacement 41

A sports club has \(30\) members: \(12\) women and \(18\) men. A six-person team is selected so that every six-person group is equally likely. 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 \(30.23\%\)

Sampling with and without replacement 42

A grade level has \(24\) students. One six-person work group is selected uniformly at random. Independently, a second six-person group is selected in the same way from all \(24\) students. The two groups are allowed to have exactly the same members. 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. - For no overlap, how many students remain outside the first group?

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 \(13.79\%\)

Sampling with and without replacement 43

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

Hints

- What kinds of three-chocolate groups satisfy the first event? - Can a group fit two different kinds from your first list? - For the second event, which fillings must be represented?

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. \(\frac{4}{17}\approx23.5\%\)

Sampling with and without replacement 44

In a school raffle game, four numbers are drawn without replacement from \(25\) numbered balls, with every four-number set equally likely. 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, \(P(X=2)=\frac{\binom{4}{2}\binom{21}{2}}{\binom{25}{4}}=\frac{126}{1265}\approx0.0996\). 3. At least three matches means exactly three or exactly four, so \(P(X\ge 3)=\frac{\binom{4}{3}\binom{21}{1}+\binom{4}{4}\binom{21}{0}}{\binom{25}{4}}=\frac{17}{2530}\approx0.00672\).

Answer

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

Sampling with and without replacement 45

An urn contains \(10\) white balls and \(10\) black balls. Three balls are selected in sequence, with each ball currently in the urn equally likely to be chosen at each draw. 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

- Can the order of the selected colors change whether exactly two balls are white? - What changes in the urn after a draw under each rule? - Does either rule keep the chance of white constant from draw to draw?

Solution

1. Without replacement, \(P(X=2)=\frac{\binom{10}{2}\binom{10}{1}}{\binom{20}{3}}=\frac{450}{1140}=\frac{15}{38}\approx0.3947\). 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. \(\frac{15}{38}\approx0.3947\) 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.

Sampling with and without replacement 46

In a city with \(25{,}000\) households, \(15\%\) get a local newspaper. A survey picks \(50\) households uniformly at random without replacement. A common rule of thumb says that a binomial approximation is often reasonable if the sample is at most \(5\%\) of the population. This is a guideline, not a guarantee. 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. - Think about why removing a few households hardly changes the subscriber proportion in a much larger city. - 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 \(14.93\%\).

Sampling with and without replacement 47

An electronics store receives a shipment of \(1200\) memory chips. Historically, \(3\%\) of the chips are defective; use this rate as the approximate defect probability for the shipment. An employee inspects a uniformly random sample without replacement. Use the rule of thumb \(\frac{n}{N}\le0.05\) to decide when a binomial approximation is reasonable. It is a guideline, not a guarantee. a) Find the largest sample size \(n\) that meets this guideline. 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 \(77.31\%\).

Sampling with and without replacement 48

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 selects \(5\) seedlings without replacement so that every five-seedling group is equally likely. 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.

Sampling with and without replacement 49

An urn contains \(20\) balls labeled \(1\) through \(20\). Three balls are drawn without replacement, with each remaining ball equally likely to be chosen next. 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

- What is the greatest number of target balls that could appear in three draws? - For a given value of \(X\), what must be true of the other drawn balls? - How does a probability distribution help you describe the average number of target balls?

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> \(E(X)=0.75\)

Sampling with and without replacement 50

Two balls are drawn from the urn, with every available ball equally likely to be selected at each draw. 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.

Sampling with and without replacement 51

Three balls are drawn in sequence from the urn, with every available ball equally likely to be chosen at each draw. 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) \(\frac{3}{10}\)

Sampling with and without replacement 52

Two balls are drawn from the urn without replacement, with every remaining ball equally likely to be selected at each draw. 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>

Sampling with and without replacement 53

Two balls are drawn from the urn, with each available ball equally likely on each draw. Determine whether drawing two balls of different colors is more likely 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.

Sampling with and without replacement 54

Two numbered balls are drawn in order from the urn without replacement. Every remaining ball is equally likely on each draw. 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

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

Sampling with and without replacement 55

Each available ball in the urn is equally likely to be drawn. 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

1. With replacement, \(P(\text{SIS})=\frac{2}{9}\cdot\frac{2}{9}\cdot\frac{2}{9}=\frac{8}{729}\). 2. Without replacement, \(P(\text{TIT})=\frac{3}{9}\cdot\frac{2}{8}\cdot\frac{2}{7}=\frac{1}{42}\). 3. 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) \(\frac{1}{28}\)

Sampling with and without replacement 56

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 \(0.04\).

Sampling with and without replacement 57

A box has \(150\) light bulbs, including \(30\) faulty bulbs. Each bulb available in the box is equally likely to be picked. Ten bulbs are tested. a) Find the probability that exactly \(2\) selected bulbs are faulty if each bulb is replaced before the next draw. b) Find the probability that at least \(1\) selected bulb is faulty if every bulb is replaced after selection. c) If the bulbs are not replaced, is the chance of selecting no faulty bulbs greater or less than with replacement? Explain.

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.

Sampling with and without replacement 58

A container holds \(15\) tickets: \(5\) winning tickets and \(10\) losing tickets. Each draw is made uniformly from the available tickets. a) Six independent draws are made with replacement. Find the probability that exactly \(2\) winning tickets are drawn. b) Six tickets are selected at once without replacement, with each six-ticket group equally likely. Find the probability that exactly \(2\) are winning tickets. c) What is the minimum number of independent 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.

Sampling with and without replacement 59

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. Four batteries are drawn uniformly from one shipment, with or without replacement as stated below. a) Approximate the probability that exactly \(1\) sampled battery is defective by modeling the selections as independent draws with replacement. b) Find the exact probability of exactly \(1\) defective battery for each shipment when sampling without replacement. c) A common rule of thumb says that a binomial distribution often gives a good approximation to sampling without replacement when the sample size is at most \(5\%\) of the population. This is a guideline, not a guarantee. 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.

Sampling with and without replacement 60

A container holds \(10\) balls numbered \(1\) through \(10\). Three balls are selected without replacement so that every possible group of three balls is equally likely. 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.

Sampling with and without replacement 61

A bag contains eight balls, all red or blue. Two balls are drawn at random without replacement, with every remaining ball equally likely at each draw. 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.

Sampling with and without replacement 62

A bag contains ten balls, including \(x\) red balls. Two balls are drawn at random without replacement, with every remaining ball equally likely at each draw. 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

\(x=4\)

Sampling with and without replacement 63

A school has \(12\) raffle tickets numbered \(1\) through \(12\). Tickets \(1\), \(2\), and \(3\) win prizes. Maya receives four tickets from the organizer. A classmate says that her chance of getting at least one winning ticket is \(\frac{41}{55}\). Describe one way the organizer could assign four tickets that makes the classmate's probability correct. Describe a second rule with a different probability. Calculate the chance of at least one win under both rules. Can Maya's chance be found from the number of tickets alone?

Hints

- What does the classmate's number assume about the possible four-ticket groups? - How else might an organizer choose which four tickets to give out? - Is there an easier opposite event to count under the fair-shuffle rule?

Solution

1. One rule shuffles all twelve tickets uniformly at random and deals Maya four of them without replacement. Then every four-ticket set is equally likely, so her chance of at least one winner is \(1-\frac{\binom{9}{4}}{\binom{12}{4}}=1-\frac{126}{495}=\frac{41}{55}\). 2. Another rule always gives Maya tickets \(1,2,3,4\). Then she certainly gets a winning ticket, so the probability is \(1\). 3. Both rules give her four tickets but different chances, so the number of tickets alone is not enough to determine the probability.

Answer

Fairly shuffled four-ticket groups give \(\frac{41}{55}\). Always receiving tickets \(1,2,3,4\) gives \(1\). The assignment rule matters.

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