Using Two-Way Tables for Probability - Statistics and Probability

1. A random sample of 170 students was asked whether they prefer morning classes or afternoon classes and whether they drink coffee. Using the two-way table, are the events A = "student drinks coffee" and B = "student prefers morning classes" independent? Compare $P(A\mid B)$ to $P(A)$.
   Answer: No. $P(A\mid B)=\frac{40}{100}=0.40$ and $P(A)=\frac{70}{170}\approx 0.412$, so they do not match. This problem tests independence of events A (drinks coffee) and B (prefers morning classes) using a two-way table. Events are independent if P(A|B) equals P(A). First, find P(A|B): given the student prefers morning classes, we examine those 100 students, and 40 drink coffee, so P(A|B) = 40/100 = 0.40. Next, find P(A): out of all 170 students, 70 drink coffee, so P(A) = 70/170 ≈ 0.412. Since 0.40 ≠ 0.412, the events are NOT independent, though they're close. Students who prefer morning classes are slightly less likely to drink coffee than the overall population. The correct answer recognizes this small but real difference. A common mistake is rounding and concluding they're "close enough" to be independent, but mathematical independence requires exact equality.
2. A random sample of 150 students was surveyed about whether they use a planner and whether they turn in homework on time. Using the two-way table, what is $P(\text{On time} \mid \text{Uses a planner})$? Give your answer as a decimal. Two-way table (counts): - Rows: Uses a planner (Yes, No) - Columns: Homework on time (Yes, No) Planner Yes: On time Yes 60, On time No 15 (Row total 75) Planner No: On time Yes 45, On time No 30 (Row total 75) Column totals: On time Yes 105, On time No 45, Grand total 150
   Answer: $\frac{60}{75}=0.80$ Conditional probabilities from two-way tables restrict to the given row or column. For P(On time | Uses a planner), 'given' is planner yes row total 75. Intersection is on time yes and planner yes, 60. So, 60/75 = 0.80. Common mistake: using column total 105, giving 60/105 ≈ 0.571, or grand total 150, 60/150 = 0.40. Circle the conditioning row (planner yes 75), find on time yes cell (60), compute 60/75. This isolates the correct denominator for the probability.
3. A random sample of 200 students was asked whether they prefer to study alone or in a group and whether they have taken an online course before. Using the two-way table, estimate the probability that a randomly selected student has taken an online course given that the student prefers to study in a group. Give your answer as a percent. Two-way table (counts): - Rows: Study preference (Alone, Group) - Columns: Taken online course (Yes, No) Alone: Yes 70, No 50 (Row total 120) Group: Yes 30, No 50 (Row total 80) Column totals: Yes 100, No 100, Grand total 200
   Answer: $\frac{30}{80}=37.5\%$ Two-way tables facilitate estimating conditional probabilities by narrowing to the given condition. Here, the 'given' is preferring to study in a group, so the denominator is the group row total of 80. The intersection, having taken an online course and preferring group study, is 30. The probability is 30/80 = 37.5%. Common errors include using the column total 100, giving 30/100 = 30%, or the grand total 200, yielding 30/200 = 15%. To compute accurately, circle the conditioning row (group total 80), identify the yes cell (30), and form 30/80. This method isolates the relevant data for the conditional estimate.
4. A cafeteria manager sampled 200 students and recorded whether each student chooses milk or juice and whether they buy breakfast at school. Using the two-way table, are the events A = "chooses milk" and B = "buys breakfast" independent? Compare $P(A\mid B)$ to $P(A)$. Two-way table (counts): - Milk & Buys breakfast: 54 - Milk & Does not buy breakfast: 66 - Juice & Buys breakfast: 36 - Juice & Does not buy breakfast: 44 Row totals: Milk 120, Juice 80 Column totals: Buys breakfast 90, Does not buy 110 Grand total: 200
   Answer: Yes; $P(A\mid B)=\frac{54}{90}$ equals $P(A)=\frac{120}{200}$. Independence in two-way tables is checked by seeing if a conditional probability equals the overall probability. For P(A|B), the 'given' is buys breakfast (90 total), with 54 choosing milk. Compute P(A|B) = 54/90 = 0.6 and P(A) = 120/200 = 0.6; equality means independent. A distractor uses the grand total as denominator for conditional, like 54/200. Strategy: circle the buys breakfast column (90), form 54/90, and compare to milk total over grand total (120/200). Matching values confirm independence.
5. A random sample of 160 students at a high school was asked what type of device they primarily use for schoolwork and whether they are in at least one school club. Using the two-way table, estimate the probability that a randomly selected student is in at least one club given that the student primarily uses a laptop. Give your answer as a decimal. Two-way table (counts): - Rows: Primary device (Laptop, Tablet, Phone) - Columns: In a club (Yes, No) Laptop: Yes 42, No 38 (Row total 80) Tablet: Yes 18, No 22 (Row total 40) Phone: Yes 12, No 28 (Row total 40) Column totals: Yes 72, No 88, Grand total 160
   Answer: $\frac{42}{80}=0.525$ Two-way tables are used to calculate conditional probabilities by focusing on specific subsets of data. The 'given' condition, here that the student primarily uses a laptop, determines the denominator as the total for that group, which is 80 students. The intersection, students in at least one club and using a laptop, is the cell with 42. Thus, the conditional probability is 42 divided by 80, equaling 0.525. A common misconception is using the grand total of 160 as the denominator instead, leading to 42/160 = 0.2625, or flipping to the column total of 72, giving 42/72 ≈ 0.5833. To avoid errors, circle the conditioning category (laptop row total 80), identify the favorable cell (42), and form the fraction 42/80. This strategy ensures you isolate the relevant subgroup for accurate estimation.
6. A random sample of 160 students at a high school was surveyed about whether they are in a school club and whether they usually bring a laptop to school. Using the two-way table, estimate the probability that a randomly selected student brings a laptop given that the student is in a school club. Give your answer as a decimal.
   Answer: $\frac{60}{90}\approx 0.667$ This problem asks for a conditional probability using a two-way table. When finding P(brings laptop | in a school club), the condition "given that the student is in a school club" tells us to focus only on students in clubs - this becomes our denominator. From the table, we need to find how many students are in clubs total (this is 90), and among those, how many bring laptops (this is 60). The conditional probability is therefore 60/90 ≈ 0.667. A common mistake is using the total sample size (160) as the denominator, which would give 60/160 = 0.375, but this ignores the conditioning. The strategy is to circle "in a school club" in the table, find that row/column total (90), then form the fraction with students who both bring laptops AND are in clubs (60) in the numerator.
7. A random sample of 160 students was surveyed about whether they have taken an art class this year (Yes/No) and whether they prefer reading or watching videos when learning something new. Using the two-way table, what is $P(\text{Art Yes}\mid\text{Reading})$? Give your answer as a fraction. Two-way table (counts): - Reading & Art Yes: 35 - Reading & Art No: 45 - Videos & Art Yes: 30 - Videos & Art No: 50 - Row totals: Reading = 80, Videos = 80 - Column totals: Art Yes = 65, Art No = 95 - Grand total: 160
   Answer: $\frac{35}{80}$ Using two-way tables for conditional probability involves selecting the 'given' group's total as the denominator. For P(Art Yes | Reading), the reading total is 80. The intersection of reading and art yes is 35. Thus, the probability is 35/80. One misconception is flipping to use the art yes total 65, yielding 35/65, but that would be P(Reading | Art Yes) instead. To compute accurately, circle the 'reading' row total 80, locate the cell 35, and form the fraction 35/80.
8. A random sample of 140 students recorded whether each student rides the bus to school and whether they arrive before the first bell. Using the two-way table, are the events A = "rides the bus" and B = "arrives before the first bell" independent? Use the table by comparing $P(A\mid B)$ to $P(A)$. Two-way table (counts): - Rows: Rides the bus (Yes, No) - Columns: Arrives before bell (Yes, No) Bus Yes: Before Yes 30, Before No 40 (Row total 70) Bus No: Before Yes 40, Before No 30 (Row total 70) Column totals: Before Yes 70, Before No 70, Grand total 140
   Answer: No; $P(A\mid B)=\frac{30}{70}\approx0.429$ does not equal $P(A)=\frac{70}{140}=0.50$. Independence in two-way tables is tested by comparing P(A|B) to P(A). The 'given' for P(A|B) is B (arrives before bell, yes total 70). Intersection A ∩ B is bus yes and before yes, 30. P(A|B) = 30/70 ≈ 0.429, P(A) = 70/140 = 0.50; unequal, so not independent. Misconception: using grand total for conditional, like 30/140 ≈ 0.214, or confusing counts like 30 vs. 40 without probabilities. Circle the conditioning column (before yes 70), get intersection (30), form 30/70, compare to 70/140. This highlights the difference, confirming dependence.
9. A student council took a random sample of 120 students and recorded whether each student participates in a school club and whether they prefer morning or afternoon announcements. Using the two-way table, estimate the probability that a randomly selected student prefers morning announcements given that the student is in a club. Give your answer as a fraction. Two-way table (counts): - In a club & Morning: 42 - In a club & Afternoon: 18 - Not in a club & Morning: 30 - Not in a club & Afternoon: 30 Row totals: In a club 60, Not in a club 60 Column totals: Morning 72, Afternoon 48 Grand total: 120
   Answer: $\frac{42}{60}$ Two-way tables are used to calculate conditional probabilities by focusing on a subset of the data. The 'given' condition, here being in a club, determines the denominator as the total number of students in that group, which is 60. The numerator is the cell where the student is in a club and prefers morning announcements, which is 42. Thus, the conditional probability is 42/60. A common misconception is using the total morning preferences (72) as the denominator instead, leading to 42/72. To avoid errors, circle the 'in a club' row total (60) and then find the fraction of that group who prefer morning (42/60). This strategy ensures you condition correctly on the given information.
10. A random sample of 180 students recorded whether they take the bus to school and whether they arrive before the first bell. Using the two-way table, what is $P(\text{arrives before bell}\mid\text{takes the bus})$? Give your answer as a decimal. Two-way table (counts): - Bus & Before bell: 64 - Bus & Not before bell: 16 - Not bus & Before bell: 70 - Not bus & Not before bell: 30 Row totals: Bus 80, Not bus 100 Column totals: Before bell 134, Not before bell 46 Grand total: 180
    Answer: $0.80$ Two-way tables allow calculation of conditional probabilities by conditioning on the given category. Here, 'given' takes the bus (80 total), with 64 arriving before the bell. The probability is 64/80 = 0.80. Misconception: using total before bell (134) as denominator, yielding 64/134 ≈ 0.48. Strategy: circle the bus row total (80), form the before-bell fraction (64/80 = 0.80). This method isolates the relevant group accurately.
11. A random sample of 120 students recorded their preferred study location (library or home) and whether they completed homework on time last week. Using the two-way table, what is $P(\text{On time}\mid\text{Library})$? Give your answer as a fraction in simplest form.
    Answer: $\frac{45}{70}=\frac{9}{14}$ This problem requires finding P(On time | Library) from a two-way table. The notation means "probability of completing homework on time given that the student studies in the library." Since we're given "Library," we only consider students who study in the library - this group becomes our denominator. Looking at the table, 70 students study in the library total, and among those, 45 completed homework on time. Therefore, P(On time | Library) = 45/70 = 9/14 in simplest form. A common error is using the total sample size (120) as the denominator, giving 45/120 = 3/8, which fails to condition on library users only. To avoid mistakes, circle the "Library" column/row, total those students (70), then place the intersection count (45) over that total.
12. A random sample of 110 students recorded whether they have a planner and whether they turned in the last project. Using the two-way table, estimate the probability that a randomly selected student turned in the last project given that the student has a planner. Give your answer as a decimal.
    Answer: $\frac{50}{70}\approx 0.714$ This problem asks for P(turned in project | has a planner) using a two-way table. The condition "given that the student has a planner" tells us to focus only on students with planners - this becomes our denominator. Looking at the table, we need to find how many students have planners total (70 students) and among those, how many turned in the last project (50 students). The conditional probability is 50/70 ≈ 0.714. A common mistake is using the total sample size (110) as denominator, giving 50/110 ≈ 0.455, which ignores the conditioning. Another error is inverting the condition. To solve correctly, circle "has a planner" in the table, find that group's total (70), then form the fraction with students who both have planners AND turned in the project (50).
13. A PE teacher randomly sampled 100 students and recorded grade level (9th or 10th) and whether the student plays a school sport. Using the two-way table, what is $P(\text{plays a sport}\mid\text{10th grade})$? Give your answer as a percent. Two-way table (counts): - 9th grade & Plays sport: 18 - 9th grade & Does not play: 32 - 10th grade & Plays sport: 30 - 10th grade & Does not play: 20 Row totals: 9th 50, 10th 50 Column totals: Plays 48, Does not play 52 Grand total: 100
    Answer: $60\%$ Conditional probabilities from two-way tables restrict to the given category for the calculation. Here, the 'given' is 10th grade, with a total of 50 students, and 30 of them play a sport as the relevant cell. The probability is 30/50 = 60%. Students might mistakenly use the total who play sports (48) as the denominator, getting 30/48 = 62.5%, which is close but incorrect. To compute accurately, circle the 10th grade row total (50), then divide the sports cell (30) by that total to form 30/50 or 60%. This focuses on the conditioned subgroup.
14. A random sample of 100 students was asked whether they prefer fiction or nonfiction and whether they visited the school library last month. Using the two-way table, estimate the probability that a randomly selected student prefers fiction given that the student visited the library last month. Give your answer as a fraction. Two-way table (counts): - Rows: Book preference (Fiction, Nonfiction) - Columns: Visited library last month (Yes, No) Fiction: Yes 28, No 32 (Row total 60) Nonfiction: Yes 12, No 28 (Row total 40) Column totals: Yes 40, No 60, Grand total 100
    Answer: $\frac{28}{40}=\frac{7}{10}$ Two-way tables enable conditional probability calculations by using the given as the base. For preferring fiction given visited library, denominator is visited yes total 40. Intersection is fiction and yes, 28. Probability: 28/40 = 7/10. Distractors might use fiction row total 60, giving 28/60 = 7/15, or grand total 100, yielding 28/100 = 7/25. Strategy: circle conditioning column (visited yes 40), locate fiction cell (28), form 28/40. This focuses on the subgroup for precise estimation.
15. A random sample of 200 students was asked whether they prefer working in groups or alone and whether they prefer math or English. Using the two-way table, estimate the probability that a randomly selected student prefers math given that the student prefers working in groups. Give your answer as a percent.
    Answer: $\frac{60}{110}\approx 55\%$ This problem asks for P(prefers math | prefers working in groups) from a two-way table. The condition "given that the student prefers working in groups" means we only consider students who prefer group work - this becomes our denominator. From the table, we need to find the total number who prefer working in groups (110 students) and among those, how many prefer math (60 students). The conditional probability is 60/110 ≈ 0.545 or about 55%. A common error is using the total sample size (200) as denominator, giving 60/200 = 30%, which ignores the conditioning on group preference. Another mistake is confusing rows and columns. The strategy is to first circle "prefers working in groups," find that total (110), then identify how many of those also prefer math (60).
16. A random sample of 90 students reported whether they prefer Android or iPhone and whether they use their phone mostly for texting or mostly for watching videos. Using the two-way table, what is $P(\text{Android}\mid\text{Videos})$? Give your answer as a decimal.
    Answer: $\frac{25}{40}=0.625$ This problem requires finding $P(\text{Android} \mid \text{Videos})$ from a two-way table. The notation means "probability of preferring Android given that the student uses phone mostly for videos." Since we're given "Videos," we only look at students who use phones for videos - this group is our denominator. From the table, 40 students use phones mostly for videos, and among those, 25 prefer Android. Therefore, $P(\text{Android} \mid \text{Videos}) = 25/40 = 0.625$. A common error is using the total sample size (90) as denominator, giving $25/90 \approx 0.278$, which fails to condition on video users. Another mistake is computing $P(\text{Videos} \mid \text{Android}) = 25/50$ instead. The key is to circle "Videos" users, count that group (40), then find how many of those prefer Android (25).
17. A random sample of 140 students was asked whether they have a library card (Yes/No) and whether they visited the library in the last month (Visited/Not visited). Using the two-way table, estimate the probability that a randomly selected student has a library card given that the student visited the library in the last month. Give your answer as a decimal. Two-way table (counts): - Visited & Card Yes: 50 - Visited & Card No: 10 - Not visited & Card Yes: 40 - Not visited & Card No: 40 - Row totals: Visited = 60, Not visited = 80 - Column totals: Card Yes = 90, Card No = 50 - Grand total: 140
    Answer: 0.83 Conditional probabilities are found in two-way tables by using the 'given' category's total as the denominator. For P(Card Yes | Visited), the visited total is 60. The cell for visited and card yes is 50. So, 50/60 ≈ 0.83. A common mistake is using the card yes total 90, giving 50/90 ≈ 0.56, but that's P(Visited | Card Yes), reversing the condition. To get it right, circle the 'visited' row total 60, divide the intersection 50 by it, yielding approximately 0.83.
18. A school library randomly sampled 90 students and recorded whether they visited the library this week and whether they prefer fiction or nonfiction. Using the two-way table, estimate the probability that a randomly selected student prefers fiction given that the student visited the library. Give your answer as a fraction. Two-way table (counts): - Visited & Fiction: 28 - Visited & Nonfiction: 12 - Did not visit & Fiction: 30 - Did not visit & Nonfiction: 20 Row totals: Visited 40, Did not visit 50 Column totals: Fiction 58, Nonfiction 32 Grand total: 90
    Answer: $\frac{28}{40}$ Conditional probabilities from two-way tables use the given group's total as the base. The 'given' is visited the library, with 40 students, and 28 preferring fiction. The probability is $\frac{28}{40}$. Common error: using total fiction (58) as denominator, getting $\frac{28}{58}$. To prevent this, circle the visited row total (40), then divide the fiction cell by it ($\frac{28}{40}$). This keeps the focus on the conditioned subset.
19. A random sample of 180 students recorded whether they prefer math or English and whether they are in honors classes (Yes/No). Using the two-way table, what is $P(\text{Honors Yes}\mid\text{Math})$? Give your answer as a percent. Two-way table (counts): - Math & Honors Yes: 36 - Math & Honors No: 54 - English & Honors Yes: 18 - English & Honors No: 72 - Row totals: Math = 90, English = 90 - Column totals: Honors Yes = 54, Honors No = 126 - Grand total: 180
    Answer: 40% Two-way tables enable calculation of conditional probabilities by conditioning on the given category. For P(Honors Yes | Math), the math total is 90 as denominator. The intersection of math and honors yes is 36. Thus, 36/90 = 0.4 or 40%. One error is using the honors yes total 54, resulting in 36/54 ≈ 67%, but that flips the conditioning. A helpful strategy: circle the 'math' row total 90, identify the cell 36, and compute 36/90 as 40%.
20. A random sample of 150 students reported whether they prefer team projects or individual projects, and whether they have taken a computer science class (Yes/No). Using the two-way table, are the events A = "student has taken computer science" and B = "student prefers team projects" independent? Use the table to justify by comparing $P(A\mid B)$ to $P(A)$. Two-way table (counts): - Team & CS Yes: 28 - Team & CS No: 42 - Individual & CS Yes: 32 - Individual & CS No: 48 - Row totals: Team = 70, Individual = 80 - Column totals: CS Yes = 60, CS No = 90 - Grand total: 150
    Answer: Yes; $P(\text{CS Yes}\mid\text{Team})=\frac{28}{70}$ equals $P(\text{CS Yes})=\frac{60}{150}$. Two-way tables test independence by comparing conditional and marginal probabilities. For P(CS Yes | Team), the team total is 70 as denominator, intersection with CS yes is 28, so 28/70 = 0.4. The marginal P(CS Yes) is 60/150 = 0.4, which equals it, so independent. A distractor could compare joints like 28/150 to 32/150, but that's not the test. Circle the 'team' total 70, compute 28/70, and match to 60/150; equality confirms independence.

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