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Relative frequency in two-way tables

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5147498
A survey of \(120\) ninth-grade students recorded whether each student plays soccer, represented by \(S\), or participates in track and field, represented by \(T\). The results are shown in the table. <table><tr><th></th><th>\(S\)</th><th>\(S^c\)</th><th>Total</th></tr><tr><th>\(T\)</th><td>\(15\)</td><td>\(25\)</td><td>\(40\)</td></tr><tr><th>\(T^c\)</th><td>\(35\)</td><td>\(45\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(50\)</td><td>\(70\)</td><td>\(120\)</td></tr></table> One student is selected at random. a) Describe the events represented by \(P(S\cap T)\) and \(P(S\cup T)\) in words. b) Find \(P(S\cap T^c)\) and \(P(S^c\cap T)\). c) Find \(P(S\cup T)\). d) Add the probabilities for “soccer only,” “track and field only,” and “both.” Which event does this sum represent?

Hints

- The intersection symbol identifies outcomes that satisfy both conditions. - Each interior table cell represents an intersection of a row event and a column event. - The union includes the three cells for exactly one sport or both sports.

Solution

1. The event \(S\cap T\) means that the student participates in both soccer and track and field. The event \(S\cup T\) means that the student participates in soccer, track and field, or both. 2. From the table, \(P(S\cap T^c)=\frac{35}{120}=\frac{7}{24}\), and \(P(S^c\cap T)=\frac{25}{120}=\frac{5}{24}\). 3. The union contains \(15+35+25=75\) students, so \(P(S\cup T)=\frac{75}{120}=\frac{5}{8}=0.625\). 4. The sum for soccer only, track and field only, and both is \(\frac{35}{120}+\frac{25}{120}+\frac{15}{120}=\frac{75}{120}=0.625\). It represents \(P(S\cup T)\).

Answer

a) \(P(S\cap T)\) is the probability that the student participates in both soccer and track and field. \(P(S\cup T)\) is the probability that the student participates in soccer, track and field, or both. b) \(P(S\cap T^c)=\frac{7}{24}\approx0.292\) \(P(S^c\cap T)=\frac{5}{24}\approx0.208\) c) \(P(S\cup T)=\frac{5}{8}=0.625\) d) The sum is \(0.625\), and it represents \(P(S\cup T)\).
5153498
A class of \(30\) students was surveyed about whether they regularly play a sport, represented by \(S\), and whether they play a musical instrument, represented by \(M\). The results are shown in the table. <table><tr><th></th><th>\(M\)</th><th>\(M^c\)</th><th>Total</th></tr><tr><th>\(S\)</th><td>\(6\)</td><td>\(14\)</td><td>\(20\)</td></tr><tr><th>\(S^c\)</th><td>\(4\)</td><td>\(6\)</td><td>\(10\)</td></tr><tr><th>Total</th><td>\(10\)</td><td>\(20\)</td><td>\(30\)</td></tr></table> Find the probability that a randomly selected student regularly plays a sport or plays a musical instrument, or both. Give the result as a fraction and as a percent.

Hints

- The word “or” includes students who satisfy either condition or both conditions. - Identify the three interior cells that belong to the union. - You can also use the complement of “neither activity.”

Solution

1. The union includes the cells for both activities, sport only, and instrument only. 2. The number of students in the union is \(6+14+4=24\). 3. Therefore, \(P(S\cup M)=\frac{24}{30}=\frac{4}{5}\). 4. Since \(\frac{4}{5}=0.80\), the percentage is \(80\%\).

Answer

\(P(S\cup M)=\frac{4}{5}=80\%\)
5386788
A two-way table classifies \(200\) transit trips by whether the trip requires a transfer, represented by \(T\), and whether it arrives on time, represented by \(O\). <table><tr><th></th><th>\(O\)</th><th>\(O^c\)</th><th>Total</th></tr><tr><th>\(T\)</th><td>\(72\)</td><td>\(28\)</td><td>\(100\)</td></tr><tr><th>\(T^c\)</th><td>\(84\)</td><td>\(16\)</td><td>\(100\)</td></tr><tr><th>Total</th><td>\(156\)</td><td>\(44\)</td><td>\(200\)</td></tr></table> Find \(P(T\cap O^c)\), \(P(T^c\cap O)\), and \(P(T\cup O)\).

Hints

- Match each event expression to the appropriate interior cell or cells. - Divide each favorable count by the total number of trips.

Solution

1. The cell for a transfer and a late arrival contains \(28\) trips, so \(P(T\cap O^c)=\frac{28}{200}=0.14\). 2. The cell for no transfer and an on-time arrival contains \(84\) trips, so \(P(T^c\cap O)=\frac{84}{200}=0.42\). 3. The union \(T\cup O\) includes the cells with counts \(72\), \(28\), and \(84\). 4. Therefore, \(P(T\cup O)=\frac{72+28+84}{200}=\frac{184}{200}=0.92\).

Answer

\(P(T\cap O^c)=0.14\) \(P(T^c\cap O)=0.42\) \(P(T\cup O)=0.92\)
5147378
A summer camp has \(150\) campers, each of whom is either a middle school student or a high school student. Of the campers, \(90\) play soccer. Middle school students make up \(40\%\) of the soccer players, and \(30\) middle school campers do not play soccer. a) Create a complete two-way table of counts for school level and soccer participation. b) What percentage of the high school campers play soccer? Round to the nearest tenth of a percent.

Hints

- Place the directly stated values in the table first. - Pay attention to the subgroup used as the base for each percentage. - Check every row and column total. - For part b, divide by the total number of high school campers.

Solution

1. The number of middle school soccer players is \(0.40\cdot90=36\). 2. The number of high school soccer players is \(90-36=54\). 3. The total number of middle school campers is \(36+30=66\), so the total number of high school campers is \(150-66=84\). 4. The number of high school campers who do not play soccer is \(84-54=30\). 5. The requested conditional percentage is \(\frac{54}{84}\cdot100\%\approx64.3\%\).

Answer

a) <table><tr><th></th><th>Plays soccer</th><th>Does not play soccer</th><th>Total</th></tr><tr><th>Middle school</th><td>\(36\)</td><td>\(30\)</td><td>\(66\)</td></tr><tr><th>High school</th><td>\(54\)</td><td>\(30\)</td><td>\(84\)</td></tr><tr><th>Total</th><td>\(90\)</td><td>\(60\)</td><td>\(150\)</td></tr></table> b) Approximately \(64.3\%\) of the high school campers play soccer.
5147388
An electronic component is produced by two different machines. Of the \(200\) components inspected, \(120\) were made by Machine A, and the rest were made by Machine B. Quality-control results show that \(5\%\) of the components from Machine A and \(10\%\) of the components from Machine B are defective. a) Organize the data in a complete two-way table of counts. b) What percentage of all inspected components are defective?

Hints

- First find how many inspected components came from Machine B. - Find the number of defective components produced by each machine separately. - Add the defective counts before finding the overall percentage.

Solution

1. Machine B produced \(200-120=80\) of the inspected components. 2. Machine A produced \(120\cdot0.05=6\) defective components, so it produced \(120-6=114\) nondefective components. 3. Machine B produced \(80\cdot0.10=8\) defective components, so it produced \(80-8=72\) nondefective components. 4. In all, \(6+8=14\) components are defective and \(114+72=186\) are nondefective. 5. The overall defective percentage is \(\frac{14}{200}\cdot100\%=7\%\).

Answer

a) <table><tr><th></th><th>Defective</th><th>Nondefective</th><th>Total</th></tr><tr><th>Machine A</th><td>\(6\)</td><td>\(114\)</td><td>\(120\)</td></tr><tr><th>Machine B</th><td>\(8\)</td><td>\(72\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(14\)</td><td>\(186\)</td><td>\(200\)</td></tr></table> b) \(7\%\) of all inspected components are defective.
5147438
A sports club has \(400\) members. A survey found that \(220\) members use the club fitness room, \(150\) members attend group exercise classes, and \(90\) members do both. Let \(F\) be the event that a randomly selected member uses the fitness room, and let \(G\) be the event that the member attends group exercise classes. a) Organize the data in a complete two-way table of counts. b) What is the probability, as a percent, that a randomly selected member does neither activity? c) Find \(P(F\cup G)\).

Hints

- Start with the count for members who do both activities. - Use row and column totals to find the other interior cells. - The union \(F\cup G\) includes everyone who does at least one activity.

Solution

1. The number who use the fitness room but do not attend group classes is \(220-90=130\). 2. The number who attend group classes but do not use the fitness room is \(150-90=60\). 3. The number who do at least one activity is \(90+130+60=280\), so the number who do neither is \(400-280=120\). 4. The probability of doing neither activity is \(\frac{120}{400}=0.30=30\%\). 5. The probability of doing at least one activity is \(P(F\cup G)=\frac{280}{400}=0.70=70\%\).

Answer

a) <table><tr><th></th><th>Attends group classes</th><th>Does not attend group classes</th><th>Total</th></tr><tr><th>Uses fitness room</th><td>\(90\)</td><td>\(130\)</td><td>\(220\)</td></tr><tr><th>Does not use fitness room</th><td>\(60\)</td><td>\(120\)</td><td>\(180\)</td></tr><tr><th>Total</th><td>\(150\)</td><td>\(250\)</td><td>\(400\)</td></tr></table> b) \(30\%\) c) \(P(F\cup G)=0.70\), or \(70\%\).
5147448
The incomplete two-way table shows relative frequencies 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></td><td></td><td>\(0.65\)</td></tr><tr><th>\(A^c\)</th><td>\(0.15\)</td><td></td><td></td></tr><tr><th>Total</th><td>\(0.40\)</td><td></td><td>\(1.00\)</td></tr></table> a) Complete the table. b) Find \(P(A^c\cap B^c)\). c) Find \(P(A\cup B)\) and briefly explain your calculation.

Hints

- All relative frequencies in the table add to \(1\). - Use a row or column total to find a missing interior value. - For the union, account for the overlap between the two events.

Solution

1. The missing row total is \(P(A^c)=1.00-0.65=0.35\), and the missing column total is \(P(B^c)=1.00-0.40=0.60\). 2. The upper-left cell is \(P(A\cap B)=0.40-0.15=0.25\). 3. The other cell in row \(A\) is \(P(A\cap B^c)=0.65-0.25=0.40\). 4. The remaining interior cell is \(P(A^c\cap B^c)=0.35-0.15=0.20\). 5. For the union, \(P(A\cup B)=0.65+0.40-0.25=0.80\). The intersection is subtracted because it was counted twice.

Answer

a) <table><tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr><tr><th>\(A\)</th><td>\(0.25\)</td><td>\(0.40\)</td><td>\(0.65\)</td></tr><tr><th>\(A^c\)</th><td>\(0.15\)</td><td>\(0.20\)</td><td>\(0.35\)</td></tr><tr><th>Total</th><td>\(0.40\)</td><td>\(0.60\)</td><td>\(1.00\)</td></tr></table> b) \(P(A^c\cap B^c)=0.20\) c) \(P(A\cup B)=0.80\). Add \(P(A)\) and \(P(B)\), then subtract \(P(A\cap B)\) to correct for double-counting.
5147458
At a school with \(250\) ninth-grade students, a survey asked who likes algebra, represented by \(A\), and band, represented by \(B\). Of the students, \(150\) like algebra, \(100\) like band, and \(60\) like both. a) Create a complete two-way table of counts. b) Is a randomly selected student more likely to like exactly one of the two subjects or neither subject? Support your answer with calculations. c) Find \(P(A\cup B)\) as a percent.

Hints

- “Exactly one” corresponds to the two off-diagonal interior cells. - The union includes students who like either subject or both. - Check that all four interior counts add to \(250\).

Solution

1. The number who like algebra but not band is \(150-60=90\). 2. The number who like band but not algebra is \(100-60=40\). 3. The number who like at least one subject is \(60+90+40=190\), so \(250-190=60\) like neither. 4. Exactly one subject is liked by \(90+40=130\) students, so its probability is \(\frac{130}{250}=0.52\). 5. The probability of liking neither subject is \(\frac{60}{250}=0.24\). Therefore, liking exactly one subject is more likely. 6. The union probability is \(P(A\cup B)=\frac{190}{250}=0.76=76\%\).

Answer

a) <table><tr><th></th><th>Likes band</th><th>Does not like band</th><th>Total</th></tr><tr><th>Likes algebra</th><td>\(60\)</td><td>\(90\)</td><td>\(150\)</td></tr><tr><th>Does not like algebra</th><td>\(40\)</td><td>\(60\)</td><td>\(100\)</td></tr><tr><th>Total</th><td>\(100\)</td><td>\(150\)</td><td>\(250\)</td></tr></table> b) A student is more likely to like exactly one subject: \(52\%\), compared with \(24\%\) for neither subject. c) \(P(A\cup B)=76\%\).
5147478
A school surveyed \(120\) students about where they get lunch. Of the students, \(70\) are in eighth grade and \(50\) are in ninth grade. A total of \(40\) students bring lunch from home, while the rest eat in the cafeteria. Of the ninth-grade students, \(15\) bring lunch from home. a) Complete a two-way table of counts for grade level and lunch source. b) What is the probability that a randomly selected eighth-grade student eats in the cafeteria? Give the result as a percent to the nearest tenth.

Hints

- Subtract the ninth-grade lunch count from the total lunch count. - Use the eighth-grade row total to find the remaining cell. - In part b, use the eighth-grade total as the denominator.

Solution

1. Of the \(40\) students who bring lunch, \(40-15=25\) are eighth-grade students. 2. The number of eighth-grade students who eat in the cafeteria is \(70-25=45\). 3. The number of ninth-grade students who eat in the cafeteria is \(50-15=35\). 4. In all, \(45+35=80\) students eat in the cafeteria. 5. Among eighth-grade students, the probability of eating in the cafeteria is \(\frac{45}{70}=\frac{9}{14}\approx0.643\), or approximately \(64.3\%\).

Answer

a) <table><tr><th></th><th>Brings lunch</th><th>Eats in cafeteria</th><th>Total</th></tr><tr><th>Eighth grade</th><td>\(25\)</td><td>\(45\)</td><td>\(70\)</td></tr><tr><th>Ninth grade</th><td>\(15\)</td><td>\(35\)</td><td>\(50\)</td></tr><tr><th>Total</th><td>\(40\)</td><td>\(80\)</td><td>\(120\)</td></tr></table> b) Approximately \(64.3\%\) of the eighth-grade students eat in the cafeteria.
5147518
A survey of \(200\) students recorded whether they use a bicycle, represented by \(B\), or public transportation, represented by \(T\), to travel to school. <table><tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr><tr><th>\(T\)</th><td>\(30\)</td><td>\(90\)</td><td>\(120\)</td></tr><tr><th>\(T^c\)</th><td>\(50\)</td><td>\(30\)</td><td>\(80\)</td></tr><tr><th>Total</th><td>\(80\)</td><td>\(120\)</td><td>\(200\)</td></tr></table> a) Find \(P(B\cup T)\). b) Compare \(P(B)+P(T)\) with \(P(B\cup T)\). Explain the difference. c) What is the probability that a student uses at most one of the two forms of transportation?

Hints

- When the two marginal probabilities are added, consider what happens to students in the overlap. - “At most one” includes everything except “both.” - Use the interior cells to check your union calculation.

Solution

1. The union includes the three cells for bicycle only, public transportation only, and both. Therefore, \(P(B\cup T)=\frac{50+90+30}{200}=\frac{170}{200}=0.85\). 2. The marginal probabilities are \(P(B)=\frac{80}{200}=0.40\) and \(P(T)=\frac{120}{200}=0.60\), so \(P(B)+P(T)=1.00\). 3. This sum exceeds \(P(B\cup T)\) by \(1.00-0.85=0.15\). The difference equals \(P(B\cap T)=\frac{30}{200}=0.15\), because students who use both forms of transportation were counted twice. 4. Using at most one form of transportation is the complement of using both. Its probability is \(1-0.15=0.85\).

Answer

a) \(P(B\cup T)=0.85\) b) \(P(B)+P(T)=1.00\), which is \(0.15\) greater than \(P(B\cup T)\). The overlap \(P(B\cap T)=0.15\) was counted twice. c) The probability is \(0.85\).
5147768
A survey of \(120\) students asked whether they play a musical instrument. Of the \(64\) eighth-grade students, \(28\) play an instrument. Altogether, \(40\) students play an instrument. a) Create a complete two-way table of counts for grade level and whether a student plays an instrument. b) How many ninth-grade students do not play an instrument? c) What percent of the ninth-grade students play an instrument? Round to the nearest tenth of a percent.

Hints

- First find the number of ninth-grade students. - Use the column total for students who play an instrument. - For part c, use the ninth-grade total as the denominator.

Solution

1. The number of ninth-grade students is \(120-64=56\). 2. The number of eighth-grade students who do not play an instrument is \(64-28=36\). 3. The number of ninth-grade students who play an instrument is \(40-28=12\). 4. The number of ninth-grade students who do not play an instrument is \(56-12=44\). 5. The percentage of ninth-grade students who play an instrument is \(\frac{12}{56}\cdot100\%\approx21.4\%\).

Answer

a) <table><tr><th></th><th>Plays an instrument</th><th>Does not play an instrument</th><th>Total</th></tr><tr><th>Eighth grade</th><td>\(28\)</td><td>\(36\)</td><td>\(64\)</td></tr><tr><th>Ninth grade</th><td>\(12\)</td><td>\(44\)</td><td>\(56\)</td></tr><tr><th>Total</th><td>\(40\)</td><td>\(80\)</td><td>\(120\)</td></tr></table> b) \(44\) ninth-grade students do not play an instrument. c) Approximately \(21.4\%\) of the ninth-grade students play an instrument.
5147778
A survey of \(500\) students asked about their reading habits. Ninth-grade students make up \(45\%\) of those surveyed. Of the eighth-grade students, \(60\%\) regularly read books. Among the ninth-grade students, \(120\) do not regularly read books. a) Create a complete two-way table of counts. b) What percent of all surveyed students regularly read books?

Hints

- Convert each percentage to a count before completing the table. - Pay attention to whether a percentage refers to all students or to one grade level. - Add the regular-reader counts from both grade levels.

Solution

1. The number of ninth-grade students is \(0.45\cdot500=225\), so the number of eighth-grade students is \(500-225=275\). 2. The number of eighth-grade students who regularly read books is \(0.60\cdot275=165\), so \(275-165=110\) do not. 3. Of the \(225\) ninth-grade students, \(225-120=105\) regularly read books. 4. In all, \(165+105=270\) students regularly read books. 5. The overall percentage is \(\frac{270}{500}\cdot100\%=54\%\).

Answer

a) <table><tr><th></th><th>Reads regularly</th><th>Does not read regularly</th><th>Total</th></tr><tr><th>Eighth grade</th><td>\(165\)</td><td>\(110\)</td><td>\(275\)</td></tr><tr><th>Ninth grade</th><td>\(105\)</td><td>\(120\)</td><td>\(225\)</td></tr><tr><th>Total</th><td>\(270\)</td><td>\(230\)</td><td>\(500\)</td></tr></table> b) \(54\%\) of all surveyed students regularly read books.
5147788
A company with \(800\) employees studies participation in its fitness program by work shift. Night-shift employees make up \(25\%\) of the workforce. Of the night-shift employees, \(20\%\) participate in the fitness program. Among the day-shift employees, \(320\) do not participate. a) Complete a two-way table of counts. b) Compare the fitness-program participation rates for night-shift and day-shift employees. Which group has the higher rate? Support your answer with calculations.

Hints

- First divide the workforce into night-shift and day-shift employees. - Complete each row using its row total. - Compare within-group percentages, not just the two participant counts.

Solution

1. The number of night-shift employees is \(0.25\cdot800=200\), so the number of day-shift employees is \(800-200=600\). 2. The number of night-shift participants is \(0.20\cdot200=40\), so \(200-40=160\) night-shift employees do not participate. 3. Since \(320\) day-shift employees do not participate, \(600-320=280\) do participate. 4. The night-shift participation rate is \(20\%\). 5. The day-shift participation rate is \(\frac{280}{600}\cdot100\%\approx46.7\%\). 6. The day-shift participation rate is higher.

Answer

a) <table><tr><th></th><th>Participates</th><th>Does not participate</th><th>Total</th></tr><tr><th>Night shift</th><td>\(40\)</td><td>\(160\)</td><td>\(200\)</td></tr><tr><th>Day shift</th><td>\(280\)</td><td>\(320\)</td><td>\(600\)</td></tr><tr><th>Total</th><td>\(320\)</td><td>\(480\)</td><td>\(800\)</td></tr></table> b) The night-shift participation rate is \(20\%\), and the day-shift rate is approximately \(46.7\%\). The day-shift rate is higher.
5152148
At a music school, \(120\) students are asked which instruments they play. Of the students, \(75\) play violin, represented by \(V\), \(50\) play flute, represented by \(F\), and \(20\) play both instruments. Create a complete two-way table of counts and find the probability that a randomly selected student plays exactly one of the two instruments.

Hints

- Subtract the overlap from each instrument total. - Use the total number of students to find the count for neither instrument. - “Exactly one” corresponds to the two off-diagonal interior cells.

Solution

1. The number who play violin but not flute is \(75-20=55\). 2. The number who play flute but not violin is \(50-20=30\). 3. The number who play neither instrument is \(120-(20+55+30)=15\). 4. Exactly one instrument is played by \(55+30=85\) students. 5. The probability is \(\frac{85}{120}=\frac{17}{24}\approx0.7083\), or approximately \(70.8\%\).

Answer

<table><tr><th></th><th>\(F\)</th><th>\(F^c\)</th><th>Total</th></tr><tr><th>\(V\)</th><td>\(20\)</td><td>\(55\)</td><td>\(75\)</td></tr><tr><th>\(V^c\)</th><td>\(30\)</td><td>\(15\)</td><td>\(45\)</td></tr><tr><th>Total</th><td>\(50\)</td><td>\(70\)</td><td>\(120\)</td></tr></table> The probability that a student plays exactly one instrument is \(\frac{17}{24}\approx70.8\%\).
5153598
A company inspects \(500\) components for two defects, \(A\) and \(B\). Of the components, \(25\) have defect \(A\), \(15\) have defect \(B\), and \(5\) have both defects. a) Create a complete two-way table of relative frequencies. b) Find the probability that a randomly selected component has exactly one of the two defects. c) Find the probability that a component has neither defect.

Hints

- First complete a table of counts, then divide each entry by \(500\). - “Exactly one defect” corresponds to two interior cells. - “Neither defect” is the cell outside both defect categories.

Solution

1. The count with defect \(A\) but not \(B\) is \(25-5=20\). 2. The count with defect \(B\) but not \(A\) is \(15-5=10\). 3. The count with neither defect is \(500-(5+20+10)=465\). 4. Divide each count by \(500\): \(\frac{5}{500}=0.01\), \(\frac{20}{500}=0.04\), \(\frac{10}{500}=0.02\), and \(\frac{465}{500}=0.93\). 5. Exactly one defect has probability \(0.04+0.02=0.06\). 6. Neither defect has probability \(0.93\).

Answer

a) <table><tr><th></th><th>\(B\)</th><th>\(B^c\)</th><th>Total</th></tr><tr><th>\(A\)</th><td>\(0.01\)</td><td>\(0.04\)</td><td>\(0.05\)</td></tr><tr><th>\(A^c\)</th><td>\(0.02\)</td><td>\(0.93\)</td><td>\(0.95\)</td></tr><tr><th>Total</th><td>\(0.03\)</td><td>\(0.97\)</td><td>\(1.00\)</td></tr></table> b) The probability is \(0.06\), or \(6\%\). c) The probability is \(0.93\), or \(93\%\).
5153608
A school cafeteria surveyed \(200\) students about food preferences. Of the students, \(80\) prefer a vegetarian option, represented by \(V\), and \(150\) like pizza, represented by \(P\). Altogether, \(190\) students either prefer a vegetarian option, like pizza, or both. a) How many students both prefer a vegetarian option and like pizza? b) Create a complete two-way table of counts. c) A student claims, “A majority of students who prefer a vegetarian option like pizza.” Determine whether the claim is correct and justify your answer.

Hints

- Use the two category totals and the union total to find the overlap. - Complete the remaining cells by subtraction. - For part c, use only the students who prefer a vegetarian option as the comparison group.

Solution

1. Use inclusion–exclusion for the union: \(190=80+150-n(V\cap P)\). 2. Solving gives \(n(V\cap P)=80+150-190=40\). 3. The count in \(V\cap P^c\) is \(80-40=40\), the count in \(V^c\cap P\) is \(150-40=110\), and the count in neither category is \(200-190=10\). 4. Among the \(80\) students who prefer a vegetarian option, \(40\) like pizza. The proportion is \(\frac{40}{80}=0.50=50\%\). 5. A majority must be more than \(50\%\), so the claim is not correct.

Answer

a) \(40\) students b) <table><tr><th></th><th>Likes pizza</th><th>Does not like pizza</th><th>Total</th></tr><tr><th>Prefers vegetarian option</th><td>\(40\)</td><td>\(40\)</td><td>\(80\)</td></tr><tr><th>Does not prefer vegetarian option</th><td>\(110\)</td><td>\(10\)</td><td>\(120\)</td></tr><tr><th>Total</th><td>\(150\)</td><td>\(50\)</td><td>\(200\)</td></tr></table> c) The claim is not correct. Exactly \(50\%\) of the students who prefer a vegetarian option like pizza, and a majority must be greater than \(50\%\).
5155728
A school surveyed \(200\) students in grades 9 and 10 about their cafeteria meal choice. There are \(90\) ninth-grade students, represented by \(G_9\), and \(110\) tenth-grade students, represented by \(G_{10}\). Of the ninth-grade students, \(60\%\) chose the vegetarian meal, represented by \(V\). Of the tenth-grade students, \(40\%\) chose the vegetarian meal. a) Create a complete two-way table of counts. b) Compare \(P(G_9\cap V)\) and \(P(G_{10}\cap V)\) for a student selected at random from the entire surveyed group. c) Let \(E=(G_9\cap V)\cup(G_{10}\cap V)\). Explain what \(E\) means in context and find \(P(E)\).

Hints

- Convert each within-grade percentage to a count. - For part b, both probabilities use all \(200\) surveyed students as the sample space. - In part c, combine the vegetarian-meal students from both grades.

Solution

1. The number of ninth-grade students who chose the vegetarian meal is \(0.60\cdot90=54\), so \(90-54=36\) did not. 2. The number of tenth-grade students who chose the vegetarian meal is \(0.40\cdot110=44\), so \(110-44=66\) did not. 3. In all, \(54+44=98\) students chose the vegetarian meal, and \(36+66=102\) did not. 4. The intersection probabilities are \(P(G_9\cap V)=\frac{54}{200}=0.27\) and \(P(G_{10}\cap V)=\frac{44}{200}=0.22\). Therefore, \(G_9\cap V\) is more likely. 5. Event \(E\) includes every surveyed student who chose the vegetarian meal, regardless of grade. Thus, \(P(E)=\frac{98}{200}=0.49\).

Answer

a) <table><tr><th></th><th>Vegetarian meal</th><th>Other meal</th><th>Total</th></tr><tr><th>Grade 9</th><td>\(54\)</td><td>\(36\)</td><td>\(90\)</td></tr><tr><th>Grade 10</th><td>\(44\)</td><td>\(66\)</td><td>\(110\)</td></tr><tr><th>Total</th><td>\(98\)</td><td>\(102\)</td><td>\(200\)</td></tr></table> b) \(P(G_9\cap V)=0.27\) is greater than \(P(G_{10}\cap V)=0.22\). c) \(E\) is the event that the selected student chose the vegetarian meal. \(P(E)=0.49\).

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