Free math worksheets

- Compare the outcomes listed in the two events. - What must the intersection look like for two events to be disjoint?

1. Event \(A\) contains only \(1\) and \(2\), while event \(B\) contains only \(5\) and \(6\). 2. The events have no outcomes in common, so \(A\cap B=\varnothing\). 3. Therefore, \(A\) and \(B\) are disjoint.

Yes. \(A\cap B=\varnothing\), so the events are disjoint.

- What happens to the overlap term in the addition rule when the events are disjoint? - The union contains either event with no double-counted outcomes.

1. Because the events are disjoint, \(P(A\cap B)=0\). 2. The addition rule therefore simplifies to \(P(A\cup B)=P(A)+P(B)=0.22+0.31=0.53\).

\(P(A\cup B)=0.53\).

- 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.

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, \(0.0188=1.88\%\).

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

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

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, \(P(\text{spade or face card})=\frac{22}{52}=\frac{11}{26}\approx 0.4231\).

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

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

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, \(P=\frac{11}{20}=0.55\).

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

- List the primes through \(20\). - List the multiples of \(4\) through \(20\). - Before adding the probabilities, check whether any outcome belongs to both events.

1. The primes through \(20\) are \(2,3,5,7,11,13,17,19\), so \(P(L)=\frac{8}{20}=0.40\). 2. The multiples of \(4\) are \(4,8,12,16,20\), so their probability is \(\frac{5}{20}=0.25\). 3. None of the multiples of \(4\) is prime, so the events are mutually exclusive. 4. Therefore, the probability of “divisible by \(4\) or prime” is \(\frac{8}{20}+\frac{5}{20}=\frac{13}{20}=0.65\).

a) \(0.40\) b) \(0.25\) c) \(0.65\); the events are mutually exclusive, so their probabilities may be added directly.

- Count the perfect squares and the multiples of \(30\) in the sample space. - Check whether the two sets overlap before adding their probabilities. - “Neither” is the complement of “at least one.”

1. There are \(14\) perfect squares from \(1^2\) through \(14^2=196\). 2. There are \(6\) multiples of \(30\): \(30,60,90,120,150,180\). 3. No number through \(200\) belongs to both groups, so the events are mutually exclusive and the union probability is \(\frac{14+6}{200}=\frac{20}{200}=0.10\). 4. The probability of neither event is the complement of the union: \(1-0.10=0.90\).

a) \(0.10\) b) \(0.90\)

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

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

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

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

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.

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

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

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

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

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

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

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

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

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, \(P(A\cup B)=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\).

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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.

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

- 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.

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 \(P(R\cap S)=\frac{45}{150}=0.30=30\%\).

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

- 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.

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 \(40\%\).

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.

- 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.

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 \(\frac{75}{120}=\frac{5}{8}=0.625=62.5\%\).

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

- 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.

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 \(400\cdot0.50=200\).

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

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

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.

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

- 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.

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 \(0.45\).

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.

- 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.

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.

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.

- “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.

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 \(\frac{25}{100}=0.25=25\%\).

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 \(25\%\).

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

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.

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.

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

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 \(\frac{12}{40}=\frac{3}{10}=0.30=30\%\).

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

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

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.

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.

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

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.

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 \(P(A\cap B)=\frac{1}{4}\ne0\).

- 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.

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.

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, \(\frac{4}{12}+\frac{5}{12}-\frac{2}{12}=\frac{7}{12}=P(A\cup B)\).

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

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 \(\frac{150}{500}=0.30=30\%\).

\(0.30=30\%\)

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

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.

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 \(P(A\cap B^c)=0.20\).

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

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 \(\frac{54}{900}=0.06=6\%\).

\(0.06=6\%\)

- 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.

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

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

- 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.

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: \(P(\text{personal or iPhone})=\frac{50}{150}+\frac{60}{150}-\frac{45}{150}=\frac{65}{150}=\frac{13}{30}\).

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

- 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?

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 \(P(S)=1-0.80=0.20\).

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

- 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.

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 \(P(C)=0.30\).

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

- 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?

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, \(P(A\cup B)\in[0.7, 1]\).

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. \(P(A\cup B)\in[0.7, 1]\)

- 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.

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 \(P(B\cap D)=0\).

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

- 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?

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 \(P(E_2\cap E_3)=0\).

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

- Check whether one component could satisfy both factory labels at once. - For mutually exclusive events, their union probability is the sum of their probabilities. - Use the table cells as counts out of the same total of \(5000\).

1. A component cannot come from both factories, so \(D_A\cap D_B=\varnothing\). 2. Therefore, \(P(D_A\cup D_B)=\frac{200}{5000}+\frac{300}{5000}=0.04+0.06=0.10\). 3. The events \(W_A\cap E\) and \(W_B\cap D\) are also mutually exclusive because they require different factories. 4. Thus, \(P((W_A\cap E)\cup(W_B\cap D))=\frac{1800}{5000}+\frac{300}{5000}=0.36+0.06=0.42\).

a) The events are mutually exclusive because one component cannot come from both factories. \(P(D_A\cup D_B)=0.10\). b) \(P((W_A\cap E)\cup(W_B\cap D))=0.42\). Direct addition is valid because the two events are mutually exclusive.

- 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?

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.

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 \(A\cap B=\emptyset\).

- 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?

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 \(P(A\cup B)\le P(A)+P(B)\).

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 \(P(A\cup B)\le P(A)+P(B)\).

- 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.”

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 \(1-P(A\cup B)=1-0.24=0.76\).

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

- 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.”

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 \(1-P(E\cup F)=1-0.85=0.15\).

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

- For each pair of divisibility events, check whether any outcomes satisfy both conditions. - Treat “neither” as the complement of a union. - “Exactly one” keeps the two nonoverlapping parts of the union and excludes the intersection.

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. Thus, the probability is \(\frac{7}{20}=0.35\). 2. There are \(6\) multiples of \(3\), \(4\) multiples of \(5\), and one common multiple, \(15\). Thus \(6+4-1=9\) outcomes are divisible by \(3\) or \(5\), so \(20-9=11\) satisfy neither condition. The probability is \(\frac{11}{20}=0.55\). 3. For divisibility by \(4\) or \(6\), the only overlap is \(12\). Exactly one condition holds for the other \(7-1=6\) outcomes in the union. Therefore, the probability is \(\frac{6}{20}=\frac{3}{10}=0.30\).

a) \(\frac{7}{20}=0.35\) b) \(\frac{11}{20}=0.55\) c) \(\frac{3}{10}=0.30\)

- 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?

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.

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.

- 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 \(E_1\).

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. \(P(E_1\cap E_2^c)=P(E_1)-P(E_1\cap E_2)=\frac{1}{6}-\frac{1}{30}=\frac{2}{15}\).

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) \(\frac{2}{15}\)

- 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.

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 \(P(R\cap E^c)=P(R)-P(E\cap R)=0.35-0.15=0.20\).

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

- 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?

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 \(P(A\cap C)=0.05\).

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

- 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.

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 \(P(S\cup M)\in[0.75,1]\).

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

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

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

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

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

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

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

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

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.

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.

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

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, \(P(A)=P(B)=0.445\).

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

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

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: \(\frac{15}{24}=\frac{5}{8}\).

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

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

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 \(0.70\).

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

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

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

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

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

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, \(n(A\cap B)=83+61-128=16\).

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

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

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 \(320-280=40\).

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

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

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.

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

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

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 \(23.3\%\).

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

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

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.

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

- 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.

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 \(P(A\cup B)=0.75\).

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 \(P(A\cup B)=0.75\).

- 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 \(B\).

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: \(P(A\cup B)=P(A)-P(A\cap B)+P(B)=P(A)+P(B)-P(A\cap B)\).

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

- 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?

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 \(P(E_1\cup E_2)\le P(E_1)+P(E_2)\).

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 \(P(E_1\cup E_2)\le P(E_1)+P(E_2)\).