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Addition rule and disjoint events

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Difficulty: 1 – 5

51475510

An electronics manufacturer inspects a sample of \(2500\) tablets for two defects: dead pixels and case damage. The inspection finds that \(35\) tablets have dead pixels, \(20\) have case damage, and \(8\) have both defects. Find the probability that a randomly selected tablet from the sample has at least one of the two defects.

Hints

- Tablets with both defects appear in each individual defect count. - Use inclusion–exclusion to avoid counting the overlap twice. - Divide the union count by the sample size.

Solution

1. Let \(D\) be the event that a tablet has dead pixels, and let \(C\) be the event that it has case damage. 2. By inclusion–exclusion, \(n(D\cup C)=35+20-8=47\). 3. Therefore, \(P(D\cup C)=\frac{47}{2500}=0.0188\). 4. As a percent, \(0.0188=1.88\%\).

Answer

\(P(D\cup C)=\frac{47}{2500}=0.0188=1.88\%\)

53081310

One card is drawn at random from a well-shuffled standard \(52\)-card deck. a) Explain why the individual cards can be treated as equally likely outcomes. b) Find the probability that the card is a spade or a face card, where face cards are jacks, queens, and kings.

Hints

- Count the spades and the face cards. - Identify the cards counted in both groups. - Use the addition rule for overlapping events.

Solution

1. After a thorough shuffle, each of the \(52\) distinct cards is modeled as equally likely to be drawn. 2. There are \(13\) spades and \(12\) face cards. The \(3\) spade face cards belong to both groups. 3. By the addition rule, the number of favorable cards is \(13+12-3=22\). Therefore, \(P(\text{spade or face card})=\frac{22}{52}=\frac{11}{26}\approx 0.4231\).

Answer

a) A thorough shuffle makes each of the \(52\) cards equally likely to be drawn. b) \(\frac{11}{26}\approx 0.4231\), or about \(42.31\%\)

53081410

A fair \(20\)-sided die is labeled with the integers \(1\) through \(20\). Find the probability that the result is prime or greater than \(15\).

Hints

- List the prime numbers through \(20\). - List the numbers greater than \(15\). - Subtract the outcomes that appear in both lists.

Solution

1. The primes from \(1\) through \(20\) are \(2,3,5,7,11,13,17,19\), giving \(8\) outcomes. 2. The numbers greater than \(15\) are \(16,17,18,19,20\), giving \(5\) outcomes. 3. The overlap is \(\{17,19\}\), with \(2\) outcomes. Therefore, the union contains \(8+5-2=11\) outcomes. 4. Since the die is fair, \(P=\frac{11}{20}=0.55\).

Answer

\(\frac{11}{20}=0.55\), or \(55\%\)

53753510

Of \(600\) website visitors, \(360\) used the app. Among the app users, \(216\) made a purchase. Among the other \(240\) visitors, \(72\) made a purchase. Find the probability that a randomly selected visitor made a purchase.

Figure for problem 537535

Hints

- Combine the purchasers from the two disjoint visitor groups. - Divide the total number of purchasers by all visitors.

Solution

1. The total number of purchasers is \(216+72=288\). 2. Therefore, the probability is \(\frac{288}{600}=0.48\).

Answer

\(0.48\), or \(48\%\)

53866210

Two diagrams are shown. Which diagram can represent the events “the die result is even” and “the die result is odd” for one roll of a fair six-sided die? Justify your choice.

Figure for problem 538662

Hints

- Can a single outcome be both even and odd? - Compare your conclusion with whether the circles overlap.

Solution

1. No die result is both even and odd. 2. Therefore, the two events have no outcomes in common and must be represented by disjoint circles. 3. Diagram b) shows disjoint circles, so it is the correct diagram.

Answer

Diagram b), because the events have no outcomes in common.

53867910

For two events, \(P(A)=0.42\), \(P(B)=0.35\), and \(P(A\cap B)=0.18\). Use the addition rule to find \(P(A\cup B)\).

Hints

- Identify the overlap that would be counted twice by simple addition. - Substitute the three given probabilities into the addition rule.

Solution

1. Apply the addition rule: \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). 2. Substitute: \(P(A\cup B)=0.42+0.35-0.18=0.59\).

Answer

\(P(A\cup B)=0.59=59\%\)

53868010

Of \(70\) recreation-center members, \(32\) use the swimming pool, \(27\) use the climbing wall, and \(12\) use both. How many members use at least one of the two facilities? Use inclusion–exclusion for counts.

Hints

- Identify the group included in both individual counts. - Correct the double-counting once.

Solution

1. Adding \(32\) and \(27\) counts the \(12\) members who use both facilities twice. 2. Therefore, \(n(A\cup B)=32+27-12=47\).

Answer

\(47\) members use at least one of the two facilities.

53868110

A fair six-sided die is rolled. Let \(A=\{1\}\) and \(B=\{6\}\). Find \(P(A\cup B)\) and explain why the addition rule simplifies in this case.

Hints

- Decide whether one die result can belong to both events. - Use the intersection probability in the addition rule.

Solution

1. The events share no outcomes, so \(P(A\cap B)=0\). 2. Therefore, the addition rule simplifies to \(P(A\cup B)=P(A)+P(B)\). 3. Thus, \(P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\).

Answer

\(P(A\cup B)=\frac{1}{3}\). The events are mutually exclusive, so there is no overlap to subtract.

53868210

For \(P(A)=0.55\), \(P(B)=0.48\), and \(P(A\cap B)=0.21\), Maya calculates \(P(A\cup B)=0.55+0.48=1.03\). Explain the error and correct the calculation.

Hints

- A probability greater than \(1\) signals an error. - Identify the region included in both individual probabilities.

Solution

1. The overlap with probability \(0.21\) is included in both individual probabilities, so it was counted twice. 2. Subtract the overlap once: \(P(A\cup B)=0.55+0.48-0.21=0.82\).

Answer

The intersection was counted twice. The correct result is \(P(A\cup B)=0.82\).

53868610

In a population, \(62\%\) have characteristic \(A\), \(47\%\) have characteristic \(B\), and \(29\%\) have both. What percent have at least one of the two characteristics?

Hints

- The overlap is counted twice when the individual percentages are added. - “At least one” refers to the union, not exactly one.

Solution

1. The two individual percentages each include the overlap. 2. Apply inclusion–exclusion: \(P(A\cup B)=62\%+47\%-29\%=80\%\).

Answer

\(80\%\) have at least one of the two characteristics.

53869110

At a community festival, \(180\) guests buy a meal wristband. Of those guests, \(96\) choose a vegetarian entree, represented by \(V\), \(74\) choose dessert, represented by \(D\), and \(38\) choose both. How many choose at least one of the two options, and how many choose neither?

Hints

- Guests who choose both options are included in both individual counts. - Find the union count before subtracting from the total.

Solution

1. Inclusion–exclusion gives \(n(V\cup D)=96+74-38=132\). 2. The number who choose neither option is \(180-132=48\).

Answer

\(132\) guests choose at least one option, and \(48\) choose neither.

53869410

A bakery offers soup, represented by \(S\), and sandwiches, represented by \(W\), at lunch. Of \(90\) orders, \(44\) include soup, \(53\) include a sandwich, and \(21\) include both. An employee claims that \(97\) orders include at least one of the two items. Evaluate and correct the claim.

Hints

- Identify the orders included in both individual counts. - Correct the double-counting before evaluating the claim.

Solution

1. The sum \(44+53=97\) counts the \(21\) orders containing both items twice. 2. Inclusion–exclusion gives \(n(S\cup W)=44+53-21=76\).

Answer

The claim is incorrect. \(76\) orders include at least one item.

53871010

A delivery company studies \(250\) packages. Let \(A\) be the event that a package is delivered on the first attempt, and let \(B\) be the event that it is delivered to a pickup locker. Of the packages, \(175\) satisfy \(A\), \(92\) satisfy \(B\), and \(54\) satisfy both. Evaluate the claim: “Exactly \(213\) packages satisfy at least one of the two conditions.”

Hints

- Substitute the three counts into inclusion–exclusion. - Compare the result with the claim.

Solution

1. Inclusion–exclusion gives \(n(A\cup B)=175+92-54=213\). 2. The calculated union count matches the stated count, so the claim is correct.

Answer

The claim is correct: \(213\) packages satisfy at least one of the two conditions.

51475210

A recreation club has \(150\) members. Of the members, \(90\) regularly run, \(75\) regularly swim, and \(30\) do neither activity. Find the probability that a randomly selected member does both activities.

Hints

- First find how many members do at least one activity. - Adding the two activity counts double-counts members who do both. - Divide the overlap count by the total membership.

Solution

1. The number who do at least one activity is \(150-30=120\). 2. Let \(R\) represent running and \(S\) represent swimming. Inclusion–exclusion gives \(120=90+75-n(R\cap S)\). 3. Therefore, \(n(R\cap S)=90+75-120=45\). 4. The probability is \(P(R\cap S)=\frac{45}{150}=0.30=30\%\).

Answer

\(P(R\cap S)=0.30=30\%\)

51475410

In a town, \(60\%\) of households read local news source \(A\), and \(50\%\) read local news source \(B\). A survey shows that \(20\%\) of households read neither source. Determine whether it is possible for exactly \(40\%\) of households to read both sources. Justify your conclusion using the addition rule for probabilities.

Hints

- First find the probability that a household reads at least one source. - Use the addition rule for the union of two events. - Solve the equation for the intersection probability.

Solution

1. Since \(20\%\) read neither source, \(P(A\cup B)=1-0.20=0.80\). 2. The addition rule gives \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). 3. Substitute the known values: \(0.80=0.60+0.50-P(A\cap B)\). 4. Solving gives \(P(A\cap B)=1.10-0.80=0.30\). 5. Therefore, the overlap must be \(30\%\), not \(40\%\).

Answer

No. The conditions require \(P(A\cap B)=0.60+0.50-0.80=0.30\), so exactly \(30\%\) of households must read both sources.

51475610

A survey of \(120\) students found that \(65\) participate in a club sport, \(50\) play a musical instrument, and \(25\) do neither activity. a) How many students do both activities? b) What is the probability that a randomly selected student does exactly one of the two activities?

Hints

- Use the count for neither activity to find the union count. - Use inclusion–exclusion to find the overlap. - “Exactly one” includes the two nonoverlapping parts outside the intersection.

Solution

1. The number who do at least one activity is \(120-25=95\). 2. Let \(S\) represent club sports and \(M\) represent playing an instrument. Inclusion–exclusion gives \(95=65+50-n(S\cap M)\). 3. Therefore, \(n(S\cap M)=65+50-95=20\). 4. The number who do exactly one activity is \((65-20)+(50-20)=45+30=75\). 5. The probability is \(\frac{75}{120}=\frac{5}{8}=0.625=62.5\%\).

Answer

a) \(20\) students b) \(\frac{5}{8}=0.625=62.5\%\)

51475710

At a bakery, \(80\%\) of customers buy bread, \(45\%\) buy cake, and \(30\%\) buy both. a) Find the probability that a customer buys neither bread nor cake. b) On a Saturday, \(400\) customers visit the bakery. Based on these rates, how many customers would be expected to buy bread but not cake?

Hints

- First find the probability of buying at least one item. - “Neither” is the complement of the union. - Bread buyers consist of bread-only buyers and buyers of both items.

Solution

1. Let \(B\) represent buying bread and \(C\) represent buying cake. 2. By the addition rule, \(P(B\cup C)=0.80+0.45-0.30=0.95\). 3. Therefore, the probability of buying neither item is \(1-0.95=0.05\). 4. The probability of buying bread but not cake is \(P(B\cap C^c)=0.80-0.30=0.50\). 5. The expected count is \(400\cdot0.50=200\).

Answer

a) \(0.05=5\%\) b) \(200\) customers

51476510

The two-way table shows counts for two events, \(A\) and \(B\). <table><tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr><tr><th>\(A\)</th><td>\(60\)</td><td>\(100\)</td><td>\(160\)</td></tr><tr><th>\(A^c\)</th><td>\(140\)</td><td>\(100\)</td><td>\(240\)</td></tr><tr><th>Total</th><td>\(200\)</td><td>\(200\)</td><td>\(400\)</td></tr></table> Use probabilities from the table to verify the addition rule \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

Hints

- Convert the relevant table counts to probabilities. - Calculate the two sides of the equation separately. - The union consists of three interior cells.

Solution

1. From the marginal totals, \(P(A)=\frac{160}{400}=0.40\) and \(P(B)=\frac{200}{400}=0.50\). 2. From the upper-left cell, \(P(A\cap B)=\frac{60}{400}=0.15\). 3. The right side of the addition rule is \(0.40+0.50-0.15=0.75\). 4. The union contains \(60+100+140=300\) outcomes, so \(P(A\cup B)=\frac{300}{400}=0.75\). 5. Both sides equal \(0.75\), so the rule is verified for these data.

Answer

\(P(A\cup B)=\frac{300}{400}=0.75\), and \(P(A)+P(B)-P(A\cap B)=0.40+0.50-0.15=0.75\). Therefore, the two sides are equal.

51476810

A school offers two after-school clubs: theater, represented by \(T\), and coding, represented by \(C\). For a randomly selected student, \(P(T)=0.30\), \(P(C)=0.25\), and \(P(T\cup C)=0.45\). a) Find the probability that the student belongs to both clubs. b) Find the probability that the student belongs to exactly one of the two clubs. c) A student claims, “Because \(0.30+0.25=0.55\), more than half of the students must belong to at least one club.” Explain the error.

Hints

- Rearrange the addition rule to solve for the intersection. - “Exactly one” excludes students in the overlap. - Consider how students in both clubs are counted in the simple sum.

Solution

1. Apply the addition rule: \(0.45=0.30+0.25-P(T\cap C)\). 2. Thus, \(P(T\cap C)=0.30+0.25-0.45=0.10\). 3. The probability of exactly one club is the union probability minus the probability of both clubs: \(0.45-0.10=0.35\). 4. The claim double-counts students who belong to both clubs. The simple sum \(0.55\) includes the \(0.10\) overlap twice, while the union probability is \(0.45\).

Answer

a) \(P(T\cap C)=0.10\) b) The probability is \(0.35\). c) Students in both clubs are counted twice in \(P(T)+P(C)\). The overlap must be subtracted to find the union.

51477010

Two events, \(A\) and \(B\), have probabilities \(P(A)=0.30\) and \(P(B)=0.50\). a) Find \(P(A\cup B)\) if \(A\) and \(B\) are mutually exclusive. b) Suppose \(P(A\cup B)=0.70\). Find \(P(A\cap B)\). c) Explain why \(P(A\cup B)=0.90\) is impossible for the given values of \(P(A)\) and \(P(B)\).

Hints

- For mutually exclusive events, the intersection probability is \(0\). - Rearrange the addition rule to find the intersection. - Determine the greatest possible union when the individual probabilities are fixed.

Solution

1. If \(A\) and \(B\) are mutually exclusive, then \(P(A\cap B)=0\), so \(P(A\cup B)=0.30+0.50=0.80\). 2. If \(P(A\cup B)=0.70\), the addition rule gives \(P(A\cap B)=0.30+0.50-0.70=0.10\). 3. The largest possible union occurs when the intersection is \(0\), giving \(P(A\cup B)=0.80\). 4. A union probability of \(0.90\) would require \(P(A\cap B)=-0.10\), which is impossible.

Answer

a) \(P(A\cup B)=0.80\) b) \(P(A\cap B)=0.10\) c) The maximum possible union probability is \(0.80\); \(0.90\) would require a negative intersection probability.

51477110

A survey of \(100\) students found that \(75\) regularly use social media, represented by \(S\), and \(40\) regularly use streaming services, represented by \(T\). Every student uses at least one of the two services. a) Find \(n(S\cap T)\). b) Verify the count form of inclusion–exclusion: \(n(S\cup T)=n(S)+n(T)-n(S\cap T)\). c) What is the probability that a randomly selected student uses streaming services but not social media?

Hints

- “Every student uses at least one” gives the union count. - Solve the inclusion–exclusion equation for the overlap. - Subtract the overlap from the streaming-service count to find streaming only.

Solution

1. Because every student uses at least one service, \(n(S\cup T)=100\). 2. Inclusion–exclusion gives \(100=75+40-n(S\cap T)\), so \(n(S\cap T)=15\). 3. Substitution verifies the rule: \(75+40-15=100\). 4. The number who use streaming services but not social media is \(40-15=25\). 5. The requested probability is \(\frac{25}{100}=0.25=25\%\).

Answer

a) \(n(S\cap T)=15\) b) \(n(S\cup T)=75+40-15=100\), so the rule is verified. c) The probability is \(0.25\), or \(25\%\).

51477210

Determine whether each statement about events in a probability experiment is true or false. Briefly justify each answer. a) If \(P(A)+P(B)>1\), then events \(A\) and \(B\) cannot be mutually exclusive. b) If \(A\cap B=\varnothing\) and \(A\cup B=S\), then \(B=A^c\). c) For any two events in a sample, \(n(A\cup B)\le n(A)+n(B)\).

Hints

- Use the fact that probabilities are at most \(1\). - Recall the definition of a complement event. - Write the inclusion–exclusion formula for counts.

Solution

1. Statement a is true. If the events were mutually exclusive, then \(P(A\cup B)=P(A)+P(B)>1\), which is impossible. 2. Statement b is true. Disjoint events whose union is the entire sample space partition the sample space, so each is the complement of the other. 3. Statement c is true. Inclusion–exclusion gives \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\). Since \(n(A\cap B)\ge0\), the union count cannot exceed the sum of the individual counts.

Answer

a) True; otherwise the union probability would exceed \(1\). b) True; the two disjoint events together contain every outcome, so \(B=A^c\). c) True; inclusion–exclusion subtracts the nonnegative overlap count.

51477910

A bag contains \(40\) slips numbered from \(1\) through \(40\). One slip is selected at random. Find the probability that its number is divisible by \(4\) or by \(10\), including numbers divisible by both.

Hints

- List or count the multiples of each divisor. - Identify the numbers that appear in both groups. - Subtract the overlap after adding the two counts.

Solution

1. There are \(40\) equally likely outcomes. 2. The multiples of \(4\) from \(1\) through \(40\) are \(4,8,12,16,20,24,28,32,36,40\), so there are \(10\). 3. The multiples of \(10\) are \(10,20,30,40\), so there are \(4\). 4. Numbers divisible by both \(4\) and \(10\) are multiples of \(20\): \(20\) and \(40\). There are \(2\). 5. By inclusion–exclusion, the favorable count is \(10+4-2=12\). 6. The probability is \(\frac{12}{40}=\frac{3}{10}=0.30=30\%\).

Answer

\(\frac{3}{10}=0.30=30\%\)

51478110

One card is drawn at random from a standard \(52\)-card deck. a) Find the probability that the card is a heart, represented by \(H\), or a face card, represented by \(F\). A face card is a jack, queen, or king. b) Explain why simply calculating \(P(H)+P(F)\) gives an incorrect result.

Hints

- Count the hearts and the face cards separately. - Identify cards that satisfy both descriptions. - Subtract the overlap after adding the two counts.

Solution

1. A standard deck has \(13\) hearts, so \(P(H)=\frac{13}{52}\). 2. There are \(3\) face cards in each of \(4\) suits, so there are \(12\) face cards and \(P(F)=\frac{12}{52}\). 3. The heart jack, queen, and king are in both events, so \(n(H\cap F)=3\). 4. By inclusion–exclusion, \(P(H\cup F)=\frac{13+12-3}{52}=\frac{22}{52}=\frac{11}{26}\). 5. The simple sum counts the three heart face cards twice.

Answer

a) \(P(H\cup F)=\frac{11}{26}\) b) The heart jack, queen, and king belong to both events, so adding the two individual probabilities counts those cards twice.

51478610

Two events, \(A\) and \(B\), have probabilities \(P(A)=\frac{3}{8}\), \(P(B)=\frac{1}{2}\), and \(P(A\cap B)=\frac{1}{4}\). a) Find \(P(A\cup B)\). b) Find \(P(A^c\cup B)\). c) Determine whether \(A\) and \(B\) are mutually exclusive. Justify your answer.

Hints

- Apply the addition rule in part a. - For part b, consider the complement of the requested union. - Mutually exclusive events have no overlap.

Solution

1. By the addition rule, \(P(A\cup B)=\frac{3}{8}+\frac{4}{8}-\frac{2}{8}=\frac{5}{8}\). 2. The complement of \(A^c\cup B\) is \(A\cap B^c\). 3. Since \(P(A\cap B^c)=P(A)-P(A\cap B)=\frac{3}{8}-\frac{2}{8}=\frac{1}{8}\), it follows that \(P(A^c\cup B)=1-\frac{1}{8}=\frac{7}{8}\). 4. Mutually exclusive events have intersection probability \(0\). Because \(P(A\cap B)=\frac{1}{4}\ne0\), the events are not mutually exclusive.

Answer

a) \(P(A\cup B)=\frac{5}{8}\) b) \(P(A^c\cup B)=\frac{7}{8}\) c) The events are not mutually exclusive because \(P(A\cap B)=\frac{1}{4}\ne0\).

51479010

A spinner labeled with the integers \(1\) through \(12\) is spun once, and all numbers are equally likely. Let \(A\) be the event “the number is a multiple of \(3\),” and let \(B\) be the event “the number is greater than \(7\).” a) Write \(A\), \(B\), \(A\cap B\), and \(A\cup B\) in set notation. b) Find \(P(A)\), \(P(B)\), and \(P(A\cap B)\). Then verify the addition rule for this experiment.

Hints

- List the outcomes satisfying each event. - Identify the outcomes shared by the two lists. - Compare the directly counted union probability with the addition-rule result.

Solution

1. The event sets are \(A=\{3,6,9,12\}\) and \(B=\{8,9,10,11,12\}\). 2. Their intersection is \(A\cap B=\{9,12\}\), and their union is \(A\cup B=\{3,6,8,9,10,11,12\}\). 3. Therefore, \(P(A)=\frac{4}{12}\), \(P(B)=\frac{5}{12}\), and \(P(A\cap B)=\frac{2}{12}\). 4. The union has \(7\) outcomes, so \(P(A\cup B)=\frac{7}{12}\). 5. The addition rule gives \(\frac{4}{12}+\frac{5}{12}-\frac{2}{12}=\frac{7}{12}\), matching the direct result.

Answer

a) \(A=\{3,6,9,12\}\), \(B=\{8,9,10,11,12\}\), \(A\cap B=\{9,12\}\), and \(A\cup B=\{3,6,8,9,10,11,12\}\) b) \(P(A)=\frac{4}{12}\), \(P(B)=\frac{5}{12}\), and \(P(A\cap B)=\frac{2}{12}\). Also, \(\frac{4}{12}+\frac{5}{12}-\frac{2}{12}=\frac{7}{12}=P(A\cup B)\).

51521510

In a survey of \(500\) students, \(80\%\) own a tablet, represented by \(T\), \(60\%\) own a laptop, represented by \(L\), and \(50\) own neither device. Find the probability that a randomly selected student owns a tablet but not a laptop.

Hints

- Convert the percentages to counts. - Use the neither count to find the union count. - Subtract the overlap from the tablet count.

Solution

1. The device counts are \(n(T)=0.80\cdot500=400\) and \(n(L)=0.60\cdot500=300\). 2. Since \(50\) students own neither device, \(n(T\cup L)=500-50=450\). 3. Inclusion–exclusion gives \(n(T\cap L)=400+300-450=250\). 4. The tablet-only count is \(n(T\cap L^c)=400-250=150\). 5. The requested probability is \(\frac{150}{500}=0.30=30\%\).

Answer

\(0.30=30\%\)

51521610

For two events, \(A\) and \(B\), suppose \(P(A)=0.35\), \(P(B)=0.55\), and \(P(A\cap B)=0.15\). a) Find \(P(A\cup B)\). b) Find \(P(A^c\cap B^c)\). c) A student claims, “The probability that \(B\) occurs without \(A\) is greater than the probability that \(A\) occurs without \(B\).” Verify or refute the claim.

Hints

- Use the addition rule to find the union. - “Neither event” is the complement of the union. - Subtract the overlap from each individual event probability.

Solution

1. By the addition rule, \(P(A\cup B)=0.35+0.55-0.15=0.75\). 2. By the complement rule, \(P(A^c\cap B^c)=1-P(A\cup B)=1-0.75=0.25\). 3. The probability that \(B\) occurs without \(A\) is \(P(A^c\cap B)=0.55-0.15=0.40\). 4. The probability that \(A\) occurs without \(B\) is \(P(A\cap B^c)=0.35-0.15=0.20\). 5. Since \(0.40>0.20\), the claim is true.

Answer

a) \(P(A\cup B)=0.75\) b) \(P(A^c\cap B^c)=0.25\) c) The claim is true because \(P(A^c\cap B)=0.40\) and \(P(A\cap B^c)=0.20\).

51536510

One three-digit positive integer is selected at random. What is the probability that the number is divisible by \(25\) or by \(40\)?

Hints

- Count all three-digit integers first. - Count the three-digit multiples of each divisor. - Use the least common multiple to count the overlap.

Solution

1. There are \(999-100+1=900\) three-digit positive integers. 2. The three-digit multiples of \(25\) run from \(4\cdot25=100\) through \(39\cdot25=975\), giving \(39-4+1=36\) numbers. 3. The three-digit multiples of \(40\) run from \(3\cdot40=120\) through \(24\cdot40=960\), giving \(24-3+1=22\) numbers. 4. A number divisible by both \(25\) and \(40\) must be divisible by \(\operatorname{lcm}(25,40)=200\). The three-digit multiples are \(200,400,600,800\), so there are \(4\). 5. Inclusion–exclusion gives \(36+22-4=54\) favorable numbers. 6. The probability is \(\frac{54}{900}=0.06=6\%\).

Answer

\(0.06=6\%\)

51536610

One integer from \(1\) through \(100\), inclusive, is selected at random. Find the probability that the number is divisible by neither \(4\) nor \(6\).

Hints

- First count numbers divisible by at least one of the two divisors. - Use the least common multiple to count the overlap. - The requested event is the complement of the union.

Solution

1. There are \(25\) multiples of \(4\) from \(1\) through \(100\). 2. There are \(16\) multiples of \(6\) because \(16\cdot6=96\) is the greatest such multiple at most \(100\). 3. Numbers divisible by both \(4\) and \(6\) are multiples of \(\operatorname{lcm}(4,6)=12\). There are \(8\) because \(8\cdot12=96\). 4. By inclusion–exclusion, \(25+16-8=33\) numbers are divisible by \(4\) or \(6\). 5. Therefore, \(100-33=67\) numbers are divisible by neither, and the probability is \(\frac{67}{100}=0.67=67\%\).

Answer

\(\frac{67}{100}=0.67=67\%\)

52287310

A company surveys \(150\) employees about their smartphones and wireless plans. Of the employees, \(40\%\) use an iPhone, and the rest use an Android phone. There are twice as many business plans as personal plans. Among employees with a business plan, \(15\%\) use an iPhone. Find the probability that a randomly selected employee has a personal plan or uses an iPhone.

Hints

- Determine the total number of employees in each phone and plan category. - A two-way table can organize the counts. - Distinguish between a percentage of all employees and a percentage of a subgroup. - For “A or B,” use the addition rule and subtract the overlap.

Solution

1. The number of iPhone users is \(0.40\cdot 150=60\), so \(90\) employees use an Android phone. 2. Let \(x\) be the number of personal plans. Then there are \(2x\) business plans, so \(x+2x=150\). Thus, there are \(50\) personal plans and \(100\) business plans. 3. Among the business-plan users, \(0.15\cdot 100=15\) use an iPhone. Therefore, \(60-15=45\) employees both use an iPhone and have a personal plan. 4. Apply the addition rule: \(P(\text{personal or iPhone})=\frac{50}{150}+\frac{60}{150}-\frac{45}{150}=\frac{65}{150}=\frac{13}{30}\).

Answer

\(P(\text{personal plan or iPhone})=\frac{13}{30}\approx 0.4333\), or about \(43.33\%\).

52718110

A prize spinner has three sectors: gold, silver, and bronze. On each spin, exactly one sector is selected, so the events \(G\), \(S\), and \(B\) are pairwise disjoint. The following probabilities are known: \(P(G\cup S)=0.35\) \(P(S\cup B)=0.85\) \(P(G\cup B)=0.80\) Find \(P(G)\), \(P(S)\), and \(P(B)\).

Hints

- How does disjointness affect the probability of a union? - Write a system of three equations for the three unknown probabilities. - What happens when you add all three equations? - How can each pair sum be used with the total probability?

Solution

1. Because the events are pairwise disjoint, \(P(G)+P(S)=0.35\), \(P(S)+P(B)=0.85\), and \(P(G)+P(B)=0.80\). 2. Add the three equations: \(2(P(G)+P(S)+P(B))=0.35+0.85+0.80=2.00\). 3. Thus \(P(G)+P(S)+P(B)=1\). 4. Subtract each given pair sum from \(1\): \(P(B)=1-0.35=0.65\), \(P(G)=1-0.85=0.15\), and \(P(S)=1-0.80=0.20\).

Answer

\(P(G)=0.15\), \(P(S)=0.20\), and \(P(B)=0.65\).

52718210

Three events \(A\), \(B\), and \(C\) are pairwise disjoint. The probability that at least one of the events occurs is \(90\%\). Event \(A\) is half as likely as event \(B\). The probability that \(B\) or \(C\) occurs is \(70\%\). Find \(P(A)\), \(P(B)\), and \(P(C)\).

Hints

- Translate “half as likely” and “at least one” into equations. - Use disjointness to write union probabilities as sums. - Substitute one equation into another to find one unknown at a time.

Solution

1. Since the events are pairwise disjoint, \(P(A)+P(B)+P(C)=0.90\). 2. The other conditions give \(P(A)=0.5P(B)\) and \(P(B)+P(C)=0.70\). 3. Substitute the pair sum into the total: \(P(A)+0.70=0.90\), so \(P(A)=0.20\). 4. Since \(0.20=0.5P(B)\), \(P(B)=0.40\). 5. Then \(0.40+P(C)=0.70\), so \(P(C)=0.30\).

Answer

\(P(A)=0.20\), \(P(B)=0.40\), and \(P(C)=0.30\).

52718310

Two events \(A\) and \(B\) have probabilities \(P(A)=0.7\) and \(P(B)=0.4\). 1. Explain why \(A\) and \(B\) cannot be disjoint. 2. Find the interval of all possible values of \(P(A\cup B)\).

Hints

- What is the greatest possible value of a probability? - Use the addition rule for two events. - How must the events be positioned for their union to be as small as possible? - Can the probability of a union be smaller than either individual probability?

Solution

1. If \(A\) and \(B\) were disjoint, then \(P(A\cup B)=0.7+0.4=1.1\), which is impossible because a probability cannot exceed \(1\). Equivalently, the addition rule gives \(P(A\cap B)\ge 0.7+0.4-1=0.1\). 2. The union is at least as likely as either event, so \(P(A\cup B)\ge \max(0.7,0.4)=0.7\). It cannot exceed \(1\). Both endpoints are attainable: the minimum occurs when \(B\subseteq A\), and the maximum occurs when \(P(A\cap B)=0.1\). Therefore, \(P(A\cup B)\in[0.7,1]\).

Answer

1. They cannot be disjoint because disjointness would give \(P(A\cup B)=1.1>1\). In fact, \(P(A\cap B)\ge 0.1\). 2. \(P(A\cup B)\in[0.7,1]\)

52718510

The events \(A\), \(B\), \(C\), and \(D\) form a partition of the sample space. It is known that \(P(A)=0.15\), \(P(B)=2P(A)\), and \(P(C)\) is \(0.10\) greater than \(P(D)\). a) Find \(P(B)\), \(P(C)\), and \(P(D)\). b) Find \(P(A\cup B\cup C)\) and \(P(B\cap D)\).

Hints

- What must the probabilities in a partition add to? - Express the relationship between \(P(C)\) and \(P(D)\) with one variable. - What is the intersection of two different events in a partition? - Write an equation with only one unknown probability.

Solution

1. \(P(B)=2\cdot 0.15=0.30\). 2. Since the events form a partition, their probabilities add to \(1\). Let \(P(D)=d\), so \(P(C)=d+0.10\). Then \(0.15+0.30+(d+0.10)+d=1\). 3. Solving \(0.55+2d=1\) gives \(d=0.225\). Therefore, \(P(D)=0.225\) and \(P(C)=0.325\). 4. Since the events are disjoint, \(P(A\cup B\cup C)=0.15+0.30+0.325=0.775\). 5. Also, \(B\cap D=\emptyset\), so \(P(B\cap D)=0\).

Answer

a) \(P(B)=0.30\), \(P(C)=0.325\), \(P(D)=0.225\) b) \(P(A\cup B\cup C)=0.775\), \(P(B\cap D)=0\)

52718610

The events \(E_1\), \(E_2\), and \(E_3\) form a partition of a sample space. Suppose \(P(E_1)=0.25\) and \(P(E_2)=\frac{2}{3}P(E_3)\). a) Find \(P(E_2)\) and \(P(E_3)\). b) If \(E_1\), \(E_2\), and \(E_3\) are the only individual outcomes, explain why this is not an equally likely outcomes model. c) Find \(P(E_1\cup E_2)\) and \(P(E_2\cap E_3)\).

Hints

- What must the probabilities of a partition add to? - Express both unknown probabilities using one variable. - What must be true of individual outcomes in an equally likely outcomes model? - What is the intersection of two different events in a partition?

Solution

1. The probabilities in a partition add to \(1\): \(0.25+\frac{2}{3}P(E_3)+P(E_3)=1\). 2. Thus \(\frac{5}{3}P(E_3)=0.75\), so \(P(E_3)=0.75\cdot\frac{3}{5}=0.45\). 3. Then \(P(E_2)=\frac{2}{3}\cdot 0.45=0.30\). 4. The individual outcomes are not equally likely because \(0.25\), \(0.30\), and \(0.45\) are different. 5. Since the events are disjoint, \(P(E_1\cup E_2)=0.25+0.30=0.55\) and \(P(E_2\cap E_3)=0\).

Answer

a) \(P(E_2)=0.30\), \(P(E_3)=0.45\) b) The outcomes are not equally likely because their probabilities are different. c) \(P(E_1\cup E_2)=0.55\), \(P(E_2\cap E_3)=0\)

53083410

A quality-control study examines \(5000\) components from two factories. <table> <thead> <tr> <th></th> <th>Nondefective (\(E\))</th> <th>Defective (\(D\))</th> </tr> </thead> <tbody> <tr> <td>Factory A (\(W_A\))</td> <td>\(1800\)</td> <td>\(200\)</td> </tr> <tr> <td>Factory B (\(W_B\))</td> <td>\(2700\)</td> <td>\(300\)</td> </tr> </tbody> </table> a) If one of the \(5000\) tested components is selected uniformly at random, find the probability that it is defective. b) A technician claims, “The probability that a future component made at Factory A is defective is exactly \(0.1\).” Evaluate this claim by distinguishing the sample relative frequency from a model probability. c) State an event in this context whose probability can be found by adding probabilities of mutually exclusive events, and calculate it.

Hints

- Add the defective counts from both factories. - Distinguish an observed sample proportion from an unknown process probability. - Choose two events that cannot happen at the same time.

Solution

1. There are \(200+300=500\) defective components, so \(P(D)=\frac{500}{5000}=0.1\). 2. For Factory A, the observed defect relative frequency is \(\frac{200}{2000}=0.1\). If the tested components are representative and the production process remains stable, this sample value estimates the future defect probability, but it does not establish that the underlying probability is exactly \(0.1\). 3. The events “defective and from Factory A” and “defective and from Factory B” are mutually exclusive. Therefore, \(P((W_A\cap D)\cup(W_B\cap D))=\frac{200}{5000}+\frac{300}{5000}=0.04+0.06=0.1\).

Answer

a) \(0.1\) b) The observed Factory A defect rate is \(0.1\). If the sample is representative and the production process remains stable, it estimates the future model probability, but it does not prove exact equality. c) Example: The component is defective and comes from Factory A or is defective and comes from Factory B. The probability is \(0.04+0.06=0.1\).

53084710

In a market survey, \(45\%\) of respondents said they regularly buy Product A, and \(35\%\) said they regularly buy Product B. 1. Find the interval \([p_{\min},p_{\max}]\) that must contain \(P(A\cup B)\), the probability that a respondent buys Product A or Product B. 2. Interpret the lower and upper endpoints in context. What relationship between the two customer groups produces each endpoint?

Hints

- How does the size of the overlap affect the size of the union? - When does a Venn diagram cover the smallest possible total region? The largest? - Use the general addition rule. - What does it mean in context if the union probability equals the sum of the individual probabilities?

Solution

1. The addition rule is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). 2. The union is smallest when the smaller event is contained in the larger event. If \(B\subseteq A\), then \(P(A\cup B)=P(A)=0.45\). 3. The union is largest when the events are disjoint. Then \(P(A\cup B)=0.45+0.35=0.80\). 4. Therefore, \(P(A\cup B)\in[0.45,0.80]\). The lower endpoint means every Product B buyer also buys Product A. The upper endpoint means no respondent buys both products.

Answer

1. \(P(A\cup B)\in[0.45,0.80]\) 2. Lower endpoint: every Product B buyer also buys Product A, so \(B\subseteq A\). Upper endpoint: the customer groups are disjoint, so \(A\cap B=\emptyset\).

53084810

For any two events \(A\) and \(B\), the following inequality holds: \(P(A)\le P(A\cup B)\le P(A)+P(B)\) 1. Justify the left inequality using the definition of a union and the probability axioms. 2. State a set-theoretic condition under which \(P(A\cup B)=P(A)+P(B)\). 3. Show that the right inequality follows from the general addition rule.

Hints

- How is a set related to its union with another set? - When can probabilities be added without subtracting an overlap? - Write the general addition rule. - What sign must an intersection probability have?

Solution

1. Since \(A\cup B=A\cup(B\setminus A)\) and the two sets on the right are disjoint, \(P(A\cup B)=P(A)+P(B\setminus A)\ge P(A)\). 2. If \(A\cap B=\emptyset\), the events are disjoint and \(P(A\cup B)=P(A)+P(B)\). More generally, equality also holds whenever \(P(A\cap B)=0\). 3. The addition rule gives \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Since \(P(A\cap B)\ge 0\), it follows that \(P(A\cup B)\le P(A)+P(B)\).

Answer

1. \(A\subseteq A\cup B\), and additivity with nonnegativity gives \(P(A)\le P(A\cup B)\). 2. A sufficient set-theoretic condition is \(A\cap B=\emptyset\). 3. Subtracting the nonnegative quantity \(P(A\cap B)\) from \(P(A)+P(B)\) gives \(P(A\cup B)\le P(A)+P(B)\).

53084910

A spinner has \(100\) equal sections labeled \(1\) through \(100\). It is spun once. Let \(A\) be the event that the number is a multiple of \(6\), and let \(B\) be the event that the number is a multiple of \(8\). a) Find \(P(A)\) and \(P(B)\). b) Find \(P(A\cap B)\). c) Use the addition rule to find the probability that the number is divisible by \(6\) or \(8\). d) Find the probability that the number is divisible by neither \(6\) nor \(8\).

Hints

- Count the numbers from \(1\) through \(100\) that satisfy each condition. - What must a number be divisible by to be a multiple of both \(6\) and \(8\)? - Use the addition rule for “A or B.” - Use a complement for “neither.”

Solution

1. There are \(16\) multiples of \(6\) and \(12\) multiples of \(8\) from \(1\) through \(100\). Thus \(P(A)=\frac{16}{100}=0.16\) and \(P(B)=\frac{12}{100}=0.12\). 2. A number divisible by both \(6\) and \(8\) must be divisible by \(\operatorname{lcm}(6,8)=24\). The four such numbers are \(24\), \(48\), \(72\), and \(96\), so \(P(A\cap B)=\frac{4}{100}=0.04\). 3. \(P(A\cup B)=0.16+0.12-0.04=0.24\). 4. The probability of neither event is \(1-P(A\cup B)=1-0.24=0.76\).

Answer

a) \(P(A)=0.16\), \(P(B)=0.12\) b) \(P(A\cap B)=0.04\) c) \(P(A\cup B)=0.24\) d) \(0.76\)

53085010

Two events \(E\) and \(F\) satisfy \(P(E)=0.65\) and \(P(F)=0.55\). a) Explain why \(E\) and \(F\) cannot be disjoint. b) Find the smallest possible value of \(P(E\cap F)\). c) Suppose \(P(E\cup F)=0.85\). Find \(P(E\cap F)\). d) For the situation in part c), find the probability that neither event occurs.

Hints

- What is the maximum possible probability of any event? - Use the general addition rule. - Compare the sum of the individual probabilities with \(1\). - “Neither” is the complement of “at least one.”

Solution

1. If the events were disjoint, then \(P(E\cup F)=0.65+0.55=1.20\), which is impossible because a probability cannot exceed \(1\). 2. Since \(P(E\cup F)=P(E)+P(F)-P(E\cap F)\le 1\), \(P(E\cap F)\ge 0.65+0.55-1=0.20\). This lower bound is attainable. 3. If \(P(E\cup F)=0.85\), then \(P(E\cap F)=0.65+0.55-0.85=0.35\). 4. The probability that neither event occurs is \(1-P(E\cup F)=1-0.85=0.15\).

Answer

a) Disjointness would give \(P(E\cup F)=1.20>1\). b) \(0.20\) c) \(P(E\cap F)=0.35\) d) \(0.15\)

53085210

A fair \(20\)-sided die numbered \(1\) through \(20\) is rolled once. a) Find the probability that the result is divisible by \(4\) or \(6\). b) Find the probability that the result is divisible by neither \(3\) nor \(5\). c) Now suppose the die is loaded. The numbers \(1\) through \(19\) are equally likely, while \(P(20)=0.10\). Find \(P(1)\).

Hints

- Check whether any results satisfy both divisibility conditions. - Treat “neither” as the complement of “at least one.” - In part c), use the fact that all outcome probabilities add to \(1\).

Solution

1. The multiples of \(4\) are \(\{4,8,12,16,20\}\), and the multiples of \(6\) are \(\{6,12,18\}\). Their intersection is \(\{12\}\), so the union contains \(5+3-1=7\) outcomes. The probability is \(\frac{7}{20}=0.35\). 2. There are \(6\) multiples of \(3\), \(4\) multiples of \(5\), and one number, \(15\), that is a multiple of both. Thus \(6+4-1=9\) outcomes are divisible by \(3\) or \(5\). The other \(20-9=11\) outcomes satisfy neither condition, so the probability is \(\frac{11}{20}=0.55\). 3. Let the common probability for each number from \(1\) through \(19\) be \(x\). Then \(19x+0.10=1\), so \(x=\frac{0.90}{19}=\frac{9}{190}\approx 0.0474\). Therefore, \(P(1)=\frac{9}{190}\approx 0.0474\).

Answer

a) \(\frac{7}{20}=0.35\) b) \(\frac{11}{20}=0.55\) c) \(P(1)=\frac{9}{190}\approx 0.0474\)

53085710

A container holds \(100\) identical balls numbered \(1\) through \(100\). One ball is selected at random. Let \(A\) be the event that the number is a multiple of \(8\), and let \(B\) be the event that the number is a multiple of \(12\). a) Find \(P(A)\) and \(P(B)\). b) Use the general addition rule to find \(P(A\cup B)\). c) Explain why an equally likely outcomes model is reasonable in this situation.

Hints

- Count the multiples of \(8\) and \(12\) from \(1\) through \(100\). - Which numbers belong to both events? - Use the general addition rule for overlapping events. - When is it reasonable to treat all individual outcomes as equally likely?

Solution

1. There are \(12\) multiples of \(8\) and \(8\) multiples of \(12\) from \(1\) through \(100\). Thus \(P(A)=\frac{12}{100}=0.12\) and \(P(B)=\frac{8}{100}=0.08\). 2. A number in both events must be a multiple of \(\operatorname{lcm}(8,12)=24\). The multiples are \(24\), \(48\), \(72\), and \(96\), so \(P(A\cap B)=\frac{4}{100}=0.04\). 3. Therefore, \(P(A\cup B)=0.12+0.08-0.04=0.16\). 4. The model is reasonable because the balls are physically identical except for their labels and the selection is random, so each ball has the same chance of being selected.

Answer

a) \(P(A)=0.12\), \(P(B)=0.08\) b) \(P(A\cup B)=0.16\) c) The balls are identical except for their labels and are selected randomly, so equal likelihood is reasonable.

53086310

A container holds \(150\) identical balls numbered \(1\) through \(150\). One ball is selected at random. Let \(E_1\) be the event that the number is divisible by \(6\), and let \(E_2\) be the event that the number is divisible by \(10\). a) Find \(P(E_1)\) and \(P(E_2)\). b) List the outcomes in \(E_1\cap E_2\) and find \(P(E_1\cap E_2)\). c) Find \(P(E_1\cup E_2)\). d) Find the probability that the number is divisible by \(6\) but not by \(10\).

Hints

- Count the multiples of each number through \(150\). - Use the least common multiple to find numbers in both events. - Apply the addition rule for the union. - For “in \(E_1\) but not \(E_2\),” subtract the overlap from \(E_1\).

Solution

1. There are \(25\) multiples of \(6\) and \(15\) multiples of \(10\), so \(P(E_1)=\frac{25}{150}=\frac{1}{6}\) and \(P(E_2)=\frac{15}{150}=\frac{1}{10}\). 2. A number in both events is a multiple of \(\operatorname{lcm}(6,10)=30\). Thus \(E_1\cap E_2=\{30,60,90,120,150\}\), and \(P(E_1\cap E_2)=\frac{5}{150}=\frac{1}{30}\). 3. \(P(E_1\cup E_2)=\frac{1}{6}+\frac{1}{10}-\frac{1}{30}=\frac{7}{30}\). 4. \(P(E_1\cap E_2^c)=P(E_1)-P(E_1\cap E_2)=\frac{1}{6}-\frac{1}{30}=\frac{2}{15}\).

Answer

a) \(P(E_1)=\frac{1}{6}\), \(P(E_2)=\frac{1}{10}\) b) \(E_1\cap E_2=\{30,60,90,120,150\}\), \(P(E_1\cap E_2)=\frac{1}{30}\) c) \(P(E_1\cup E_2)=\frac{7}{30}\) d) \(\frac{2}{15}\)

53086410

A spinner has \(20\) equal sections labeled \(1\) through \(20\). a) Find the theoretical probability \(P(L)\) that the result is prime. b) In \(500\) spins, a prime occurred \(195\) times. Find the relative frequency and its absolute difference from \(P(L)\). c) Find the probability that the result is divisible by \(4\) or prime. Explain why the event probabilities may be added directly.

Hints

- List the primes through \(20\). - Divide the prime count by \(500\). - Check whether any multiple of \(4\) is prime.

Solution

1. The primes through \(20\) are \(2,3,5,7,11,13,17,19\), so \(P(L)=\frac{8}{20}=0.4\). 2. The relative frequency is \(\frac{195}{500}=0.39\). The absolute difference is \(|0.39-0.4|=0.01\). 3. The numbers divisible by \(4\) are \(4,8,12,16,20\). None is prime, so the two events are mutually exclusive. Therefore, \(P=\frac{8}{20}+\frac{5}{20}=\frac{13}{20}=0.65\).

Answer

a) \(0.4\) b) \(0.39\); absolute difference \(0.01\) c) \(0.65\); the events are mutually exclusive.

53086910

In a survey of \(500\) people, \(42\%\) said they exercise regularly, and \(35\%\) said they read books regularly. The relative frequency of people who do neither activity is \(0.38\). One respondent is selected at random. Let \(E\) be the event that the person exercises regularly, and let \(R\) be the event that the person reads books regularly. a) Find the probability that the person both exercises and reads books. b) Find the probability that the person reads books but does not exercise.

Hints

- Use the “neither” group to find the probability of at least one activity. - Which formula connects the two individual probabilities, their union, and their intersection? - Subtract the overlap from the group that reads books. - A Venn diagram or two-way table may help organize the regions.

Solution

1. The probability of doing at least one activity is \(1-0.38=0.62\). 2. By the addition rule, \(P(E\cap R)=P(E)+P(R)-P(E\cup R)=0.42+0.35-0.62=0.15\). 3. The probability of reading but not exercising is \(P(R\cap E^c)=P(R)-P(E\cap R)=0.35-0.15=0.20\).

Answer

a) \(P(E\cap R)=0.15\) b) \(P(R\cap E^c)=0.20\)

53087110

A container holds \(100\) tickets numbered \(1\) through \(100\). One ticket is selected at random. Let \(A\) be the event that the number is divisible by \(5\), \(B\) the event that it is divisible by \(8\), and \(C\) the event that it is divisible by \(20\). Find \(P(A)\), \(P(B)\), and \(P(C)\). Then find \(P(A\cup B)\) and \(P(A\cap C)\).

Hints

- Count the numbers from \(1\) through \(100\) that satisfy each condition. - Which numbers are in both \(A\) and \(B\)? - Use the addition rule for “A or B.” - How are the conditions “divisible by \(5\)” and “divisible by \(20\)” related?

Solution

1. There are \(20\) multiples of \(5\), \(12\) multiples of \(8\), and \(5\) multiples of \(20\) from \(1\) through \(100\). Thus \(P(A)=0.20\), \(P(B)=0.12\), and \(P(C)=0.05\). 2. The numbers divisible by both \(5\) and \(8\) are multiples of \(40\): \(40\) and \(80\). Therefore, \(P(A\cap B)=0.02\). 3. By the addition rule, \(P(A\cup B)=0.20+0.12-0.02=0.30\). 4. Every multiple of \(20\) is a multiple of \(5\), so \(C\subseteq A\). Therefore, \(A\cap C=C\), and \(P(A\cap C)=0.05\).

Answer

\(P(A)=0.20\), \(P(B)=0.12\), \(P(C)=0.05\), \(P(A\cup B)=0.30\), \(P(A\cap C)=0.05\)

53087310

In a survey, \(75\%\) of respondents use a video-streaming service regularly, and \(65\%\) have a music-app subscription. One respondent is selected at random. Find the smallest and largest possible probabilities that the respondent 1. uses both services; 2. uses at least one of the services.

Hints

- In an extreme case, could the smaller group be entirely inside the larger group? - How small can the overlap be without making the union exceed \(100\%\)? - A probability cannot exceed \(1\). - Use the addition rule to connect the union and intersection.

Solution

1. Let \(S\) be video streaming and \(M\) be a music subscription. The largest possible intersection is the smaller marginal probability: \(P(S\cap M)_{\max}=\min(0.75,0.65)=0.65\). 2. Since \(P(S\cup M)\le 1\), the intersection satisfies \(P(S\cap M)\ge 0.75+0.65-1=0.40\). Thus \(P(S\cap M)\in[0.40,0.65]\). 3. The union is smallest when the intersection is largest: \(0.75+0.65-0.65=0.75\). 4. The union is largest when the intersection is smallest: \(0.75+0.65-0.40=1\). Thus \(P(S\cup M)\in[0.75,1]\).

Answer

1. Both services: minimum \(40\%\), maximum \(65\%\) 2. At least one service: minimum \(75\%\), maximum \(100\%\)

53094410

A computer generates a uniformly random integer from \(1\) through \(200\). a) Find the probability that the number is a perfect square or a multiple of \(30\). b) In \(500\) trials, this event occurred \(42\) times. Find the relative frequency, compare it with the theoretical probability, and find the absolute difference between the observed count and the expected count.

Hints

- List the perfect squares and multiples of \(30\) through \(200\). - Check whether the two sets overlap. - Multiply the theoretical probability by \(500\) for the expected count.

Solution

1. There are \(14\) perfect squares from \(1^2\) through \(14^2=196\). There are \(6\) multiples of \(30\): \(30,60,90,120,150,180\). No number through \(200\) belongs to both groups, so \(P=\frac{14+6}{200}=0.1\). 2. The relative frequency is \(\frac{42}{500}=0.084\), which is \(0.016\) below the theoretical probability. The expected count is \(500\cdot0.1=50\), so the absolute count difference is \(|42-50|=8\).

Answer

a) \(0.1\) b) Relative frequency \(0.084\); probability difference \(0.016\); count difference \(8\)

53868310

Suppose \(P(A)=0.46\), \(P(B)=0.39\), and \(P(A\cup B)=0.68\). Find \(P(A\cap B)\).

Hints

- Rearrange the addition rule to isolate the overlap. - Check that the intersection probability is no greater than either individual probability.

Solution

1. Rearrange the addition rule: \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\). 2. Substitute: \(P(A\cap B)=0.46+0.39-0.68=0.17\).

Answer

\(P(A\cap B)=0.17=17\%\)

53868410

Suppose \(P(A)=0.28\), \(P(A\cap B)=0.11\), and \(P(A\cup B)=0.64\). Find \(P(B)\).

Hints

- Identify the known union, individual probability, and intersection. - Isolate the requested probability before substituting.

Solution

1. Rearrange the addition rule: \(P(B)=P(A\cup B)-P(A)+P(A\cap B)\). 2. Substitute: \(P(B)=0.64-0.28+0.11=0.47\).

Answer

\(P(B)=0.47=47\%\)

53868510

In two situations, \(P(A)=0.50\) and \(P(B)=0.40\). In Situation I, \(P(A\cap B)=0.10\). In Situation II, \(P(A\cap B)=0.30\). Compare \(P(A\cup B)\) in the two situations and explain the difference.

Hints

- Apply the same addition rule in both situations. - Compare only the effect of the different intersection probabilities.

Solution

1. In Situation I, \(P(A\cup B)=0.50+0.40-0.10=0.80\). 2. In Situation II, \(P(A\cup B)=0.50+0.40-0.30=0.60\). 3. With the individual probabilities fixed, a larger overlap produces a smaller union because a larger amount is subtracted in the addition rule.

Answer

Situation I: \(P(A\cup B)=0.80\) Situation II: \(P(A\cup B)=0.60\) The larger intersection in Situation II makes its union probability smaller.

53868710

For two events, \(P(A\cup B)=0.73\), \(P(A\cap B)=0.16\), and \(P(A)=P(B)\). Find both individual probabilities.

Hints

- Represent the equal unknown probabilities with one variable. - Substitute into the addition rule and solve the resulting equation.

Solution

1. Let \(P(A)=P(B)=p\). 2. The addition rule gives \(0.73=p+p-0.16\). 3. Therefore, \(2p=0.89\), so \(p=0.445\). 4. Thus, \(P(A)=P(B)=0.445\).

Answer

\(P(A)=P(B)=0.445=44.5\%\)

53868810

Suppose \(P(A)=\frac{3}{8}\), \(P(B)=\frac{5}{12}\), and \(P(A\cap B)=\frac{1}{6}\). Find \(P(A\cup B)\) as a fraction in simplest form.

Hints

- Rewrite all three fractions with a common denominator. - Simplify after combining the fractions.

Solution

1. Apply the addition rule: \(P(A\cup B)=\frac{3}{8}+\frac{5}{12}-\frac{1}{6}\). 2. Using a common denominator of \(24\), \(P(A\cup B)=\frac{9}{24}+\frac{10}{24}-\frac{4}{24}=\frac{15}{24}\). 3. Simplify: \(\frac{15}{24}=\frac{5}{8}\).

Answer

\(P(A\cup B)=\frac{5}{8}\)

53868910

Suppose \(P(A)=0.36\), \(P(B)=0.27\), and \(P(A\cup B)=0.70\). Use the addition rule to determine whether these values are possible.

Hints

- Calculate the intersection probability implied by the three values. - Check whether the result is a valid probability.

Solution

1. The addition rule would require \(P(A\cap B)=0.36+0.27-0.70=-0.07\). 2. A probability cannot be negative, so the values are impossible. 3. Equivalently, even with no overlap, the largest possible union is \(0.36+0.27=0.63\), which is less than \(0.70\).

Answer

The values are impossible because they would require \(P(A\cap B)=-0.07\).

53869210

A library analyzes \(240\) reservations. Let \(A\) be the event that an item is picked up the same day, and let \(E\) be the event that the patron also borrows an e-book. Suppose \(P(A)=0.45\), \(P(E)=0.35\), and \(P(A\cap E)=0.20\). Find \(P(A\cup E)\) and the corresponding number of reservations.

Hints

- First find the probability of at least one event. - Multiply the probability by the total number of reservations.

Solution

1. Apply the addition rule: \(P(A\cup E)=0.45+0.35-0.20=0.60\). 2. The corresponding count is \(0.60\cdot240=144\).

Answer

\(P(A\cup E)=0.60\), corresponding to \(144\) reservations.

53870110

A community pool has \(150\) guests. Of the guests, \(83\) use the lap pool, \(61\) use the water slide, and \(22\) use neither. How many guests use both?

Hints

- First find the number of guests in the union. - Compare the union count with the sum of the two individual counts.

Solution

1. The number who use at least one facility is \(150-22=128\). 2. Inclusion–exclusion gives \(128=83+61-n(A\cap B)\). 3. Therefore, \(n(A\cap B)=83+61-128=16\).

Answer

\(16\) guests use both the lap pool and the water slide.

53870210

At an evening museum event, \(210\) people visit an exhibit, \(165\) join a guided tour, and \(95\) do both. A total of \(320\) people attend. Determine whether the data are consistent, and find how many people do neither activity.

Hints

- Use inclusion–exclusion to find the union count. - Compare the union count with the total attendance, then find the remainder.

Solution

1. The number who do at least one activity is \(210+165-95=280\). 2. Since \(280\le320\), the data are possible. 3. The number who do neither activity is \(320-280=40\).

Answer

The data are consistent, and \(40\) people do neither activity.

53870310

In a hiking challenge, let \(A\) be the event that a route is longer than \(6\,\text{mi}\), and let \(B\) be the event that it has more than \(1500\,\text{ft}\) of elevation gain. For a randomly selected route, \(P(A)=0.52\), \(P(B)=0.36\), and \(P(A\cup B)=0.71\). Find \(P(A\cap B)\) and interpret the result.

Hints

- Rearrange the addition rule to find the common portion. - Translate the intersection probability back into the hiking context.

Solution

1. Rearrange the addition rule: \(P(A\cap B)=P(A)+P(B)-P(A\cup B)\). 2. Substitute: \(P(A\cap B)=0.52+0.36-0.71=0.17\). 3. Thus, \(17\%\) of the routes satisfy both conditions.

Answer

\(P(A\cap B)=0.17\). Therefore, \(17\%\) of the routes are longer than \(6\,\text{mi}\) and have more than \(1500\,\text{ft}\) of elevation gain.

53870610

At a flea market, \(120\) booths sell books, clothing, or other items. Of the booths, \(54\) sell books, \(67\) sell clothing, and \(29\) sell both. Find the probability that a randomly selected booth sells neither books nor clothing.

Hints

- First find the number of booths in the union. - Divide the remaining count by the total number of booths.

Solution

1. The number that sell books or clothing is \(54+67-29=92\). 2. The number that sell neither is \(120-92=28\). 3. The probability is \(\frac{28}{120}=\frac{7}{30}\approx0.2333\), or approximately \(23.3\%\).

Answer

\(\frac{7}{30}\approx23.3\%\)

53871110

A volunteer program asks \(84\) people whether they can work a morning shift, represented by \(A\), or an afternoon shift, represented by \(B\). Of the volunteers, \(39\) can work the morning shift, \(46\) can work the afternoon shift, and \(15\) can work both. Are there enough commitments if at least \(75\) different people must work at least one shift? Justify your answer.

Hints

- Count different people, not total shift selections. - Compare the union count with the required number.

Solution

1. The number of different people who can work at least one shift is \(n(A\cup B)=39+46-15=70\). 2. The program needs at least \(75\) people. 3. Since \(75-70=5\), the program is short by \(5\) commitments.

Answer

No. Only \(70\) different people can work at least one shift, so \(5\) more commitments are needed.

51476910

Two events, \(A\) and \(B\), have probabilities \(P(A)=0.75\) and \(P(B)=0.40\). a) Explain mathematically why \(A\) and \(B\) cannot be mutually exclusive. b) Find the smallest possible value of \(P(A\cap B)\). c) Suppose \(B\subseteq A\). Find \(P(A\cap B)\) and \(P(A\cup B)\) in this case.

Hints

- A union probability cannot be greater than \(1\). - Apply the addition rule with the largest possible value of \(P(A\cup B)\). - When one event is contained in another, identify their intersection and union.

Solution

1. If the events were mutually exclusive, then \(P(A\cup B)=0.75+0.40=1.15\), which is impossible because a probability cannot exceed \(1\). 2. The addition rule and \(P(A\cup B)\le1\) give \(0.75+0.40-P(A\cap B)\le1\). 3. Therefore, \(P(A\cap B)\ge0.15\), so the smallest possible value is \(0.15\). 4. If \(B\subseteq A\), then \(A\cap B=B\) and \(A\cup B=A\). 5. Thus, \(P(A\cap B)=0.40\) and \(P(A\cup B)=0.75\).

Answer

a) They cannot be mutually exclusive because \(P(A)+P(B)=1.15>1\). b) The smallest possible value is \(P(A\cap B)=0.15\). c) \(P(A\cap B)=0.40\) and \(P(A\cup B)=0.75\).

52717810

The probability axioms state additivity only for disjoint events. a) Use the axioms to show that for any events \(A\) and \(B\), \(P(A)=P(A\cap B)+P(A\cap B^c)\). b) Use the decomposition \(A\cup B=(A\cap B^c)\cup B\) to derive the general addition rule \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\).

Hints

- Split \(A\) into the part inside \(B\) and the part outside \(B\). - Why are \(A\cap B\) and \(A\cap B^c\) disjoint? - Solve the equation from part a) for \(P(A\cap B^c)\). - Use a disjoint decomposition of \(A\cup B\) that includes all of \(B\).

Solution

1. Split \(A\) into two disjoint parts: \(A=(A\cap B)\cup(A\cap B^c)\). 2. By additivity, \(P(A)=P(A\cap B)+P(A\cap B^c)\). 3. Also, \(A\cup B=(A\cap B^c)\cup B\), and these two events are disjoint. Therefore, \(P(A\cup B)=P(A\cap B^c)+P(B)\). 4. From part a), \(P(A\cap B^c)=P(A)-P(A\cap B)\). 5. Substitute this expression: \(P(A\cup B)=P(A)-P(A\cap B)+P(B)=P(A)+P(B)-P(A\cap B)\).

Answer

a) \(P(A)=P(A\cap B)+P(A\cap B^c)\) b) \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\)

52718410

Let \(E_1\) and \(E_2\) be any two events. Derive the inequality \(P(E_1\cup E_2)\le P(E_1)+P(E_2)\) using the probability axiom for disjoint events. Hint: Split \(E_1\cup E_2\) into two disjoint parts, one of which is \(E_1\).

Hints

- Recall the probability rule for the union of disjoint events. - Separate the overlap from the rest of \(E_2\). - How does the probability of a subset compare with the probability of the entire event?

Solution

1. Write the union as the disjoint union \(E_1\cup E_2=E_1\cup(E_2\setminus E_1)\). 2. By additivity, \(P(E_1\cup E_2)=P(E_1)+P(E_2\setminus E_1)\). 3. Also, \(E_2=(E_2\setminus E_1)\cup(E_1\cap E_2)\), and these two parts are disjoint. Therefore, \(P(E_2)=P(E_2\setminus E_1)+P(E_1\cap E_2)\ge P(E_2\setminus E_1)\). 4. Substitute this inequality into step 2 to obtain \(P(E_1\cup E_2)\le P(E_1)+P(E_2)\).

Answer

Because \(E_1\cup E_2=E_1\cup(E_2\setminus E_1)\) is a disjoint union, \(P(E_1\cup E_2)=P(E_1)+P(E_2\setminus E_1)\). Since \(P(E_2\setminus E_1)\le P(E_2)\), it follows that \(P(E_1\cup E_2)\le P(E_1)+P(E_2)\).

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