Statistics Quiz: Using Two Way Tables For Probability

What this quiz covers

This quiz focuses on Using Two Way Tables For Probability, giving you a quick way to practice the rules, question types, and explanations that matter most for Statistics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. $\frac{18}{63}$
2. $\frac{18}{60}$ (correct answer)
3. $\frac{63}{60}$
4. $\frac{18}{150}$

Explanation: Two-way tables help compute conditional probabilities like $P(\text{Morning} \mid \text{Group})$ by restricting to the 'group' category. The 'given' group row total is 60, which becomes the denominator. The intersection of group and morning is 18, so that's the numerator. The probability is $\frac{18}{60}$. One distractor might flip to use the column total 63 as denominator, giving $\frac{18}{63}$, but that's incorrect as it ignores the conditioning on group. A good strategy is to circle the conditioning row 'group' total of 60, identify the relevant cell 18, and compute the fraction $\frac{18}{60}$.

Question 2

1. 45%
2. 30% (correct answer)
3. 50%
4. 15%

Explanation: Conditional probabilities from two-way tables use the 'given' category's total as the denominator. Here, for P(sport | job yes), the job yes total is 90. The cell for job yes and sport yes is 27. So, 27/90 = 0.3 or 30%. A common error is using the sport yes column total 81 as denominator, giving 27/81 = about 33%, but that's not conditioning on job. Strategically, circle the 'job yes' row total 90, find the intersection 27, and divide to get the probability as 30%.

Question 3

1. $\frac{30}{48}$ (correct answer)
2. $\frac{30}{70}$
3. $\frac{48}{70}$
4. $\frac{30}{96}$

Explanation: In two-way tables, conditional probability uses the 'given' group's total for the denominator. For P(On time | Bus), the bus total is 48. The cell for bus and on time is 30. So, the probability is 30/48. A misconception might be using the on time total 70, giving 30/70, but that conditions on on time instead. To avoid this, circle the 'bus' row total 48, find the intersection 30, and form the fraction 30/48.

Question 4

1. 20%
2. 30%
3. 40% (correct answer)
4. 60%

Explanation: 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%.

Question 5

1. 0.36
2. 0.83 (correct answer)
3. 0.50
4. 0.71

Explanation: 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.

Question 6

1. $\frac{35}{80}$ (correct answer)
2. $\frac{35}{65}$
3. $\frac{65}{80}$
4. $\frac{35}{160}$

Explanation: 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.

Question 7

1. No; $P(\text{CS Yes}\mid\text{Team})=\frac{28}{70}$ does not equal $P(\text{CS Yes})=\frac{60}{150}$.
2. Yes; because 70 and 80 are close, the events are independent.
3. Yes; $P(\text{CS Yes}\mid\text{Team})=\frac{28}{70}$ equals $P(\text{CS Yes})=\frac{60}{150}$. (correct answer)
4. No; because $\frac{28}{150}$ is not equal to $\frac{32}{150}$.

Explanation: 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.

Question 8

1. $64\%$ (correct answer)
2. $41.6\%$
3. $29\%$
4. $70\%$

Explanation: Conditional probabilities in two-way tables are found by dividing within the given category. The 'given' is attends tutoring (50 total), with 32 passing the quiz. The probability is 32/50 = 64%. Error: using total passed (77) as base, getting 32/77 ≈ 41.6%. Circle tutoring row (50), divide passed by it (32/50 = 64%). This ensures correct conditioning.

Question 9

1. Yes; $P(A\mid B)=\frac{54}{90}$ equals $P(A)=\frac{120}{200}$. (correct answer)
2. No; $P(A\mid B)=\frac{54}{90}$ is not equal to $P(A)=\frac{120}{200}$.
3. No; $P(A\mid B)=\frac{54}{200}$ is not equal to $P(A)=\frac{120}{200}$.
4. Yes; because 54 and 66 are close.

Explanation: 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.

Question 10

1. $\frac{72}{120}$
2. $\frac{42}{120}$
3. $\frac{42}{60}$ (correct answer)
4. $\frac{42}{72}$

Explanation: 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.

Question 11

1. Yes. $P(A\mid B)=\frac{100}{170}\approx 0.588$ and $P(A)=\frac{70}{170}\approx 0.412$, so they match.
2. Yes. $P(A\mid B)=\frac{40}{100}=0.40$ and $P(A)=\frac{70}{170}\approx 0.412$, so they match.
3. No. $P(A\mid B)=\frac{40}{100}=0.40$ and $P(A)=\frac{70}{170}\approx 0.412$, so they do not match. (correct answer)
4. No. $P(A\mid B)=\frac{40}{170}\approx 0.235$ and $P(A)=\frac{70}{170}\approx 0.412$, so they do not match.

Explanation: 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.

Question 12

1. Yes. $P(A\mid B)=\frac{30}{80}=0.375$ and $P(A)=\frac{60}{140}\approx 0.429$, so they match.
2. Yes. $P(A\mid B)=\frac{30}{140}\approx 0.214$ and $P(A)=\frac{60}{140}\approx 0.429$, so they match.
3. No. $P(A\mid B)=\frac{30}{80}=0.375$ and $P(A)=\frac{60}{140}\approx 0.429$, so they do not match. (correct answer)
4. No. $P(A\mid B)=\frac{80}{140}\approx 0.571$ and $P(A)=\frac{60}{140}\approx 0.429$, so they do not match.

Explanation: This problem tests whether events A (prefers digital notes) and B (feels prepared) are independent using a two-way table. Events are independent if P(A|B) = P(A). First, find P(A|B): given the student feels prepared, we look at those 80 students, and 30 prefer digital notes, so P(A|B) = 30/80 = 0.375. Next, find P(A): out of all 140 students, 60 prefer digital notes, so P(A) = 60/140 ≈ 0.429. Since 0.375 ≠ 0.429, the events are NOT independent - students who feel prepared are less likely to prefer digital notes than the overall population. A common error is concluding independence when the probabilities are "close" rather than exactly equal. The strategy is to compute both probabilities carefully and compare - any difference means the events are dependent.

Question 13

1. $\frac{60}{90}\approx 0.667$ (correct answer)
2. $\frac{90}{160}=0.5625$
3. $\frac{60}{70}\approx 0.857$
4. $\frac{60}{160}=0.375$

Explanation: 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.

Question 14

1. $\frac{120}{180}=0.667$
2. $\frac{90}{60}=1.50$
3. $\frac{90}{120}=0.75$ (correct answer)
4. $\frac{90}{180}=0.50$

Explanation: This problem requires finding P(On time | Eats breakfast) using a two-way table. The condition "given eats breakfast" tells us to focus only on students who eat breakfast - this group forms our denominator. Looking at the table, we need to identify how many students eat breakfast total (120 students) and among those, how many are on time to first period (90 students). Therefore, P(On time | Eats breakfast) = 90/120 = 0.75. A common mistake is using the total sample size (180) as denominator, which would give 90/180 = 0.50, but this fails to condition properly. Another error is computing the inverse probability. To solve correctly, circle "Eats breakfast" in the table, find that group's total (120), then form the fraction with students who both eat breakfast AND are on time (90).

Question 15

1. $\frac{50}{110}\approx 0.455$
2. $\frac{50}{70}\approx 0.714$ (correct answer)
3. $\frac{50}{60}\approx 0.833$
4. $\frac{70}{110}\approx 0.636$

Explanation: 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).

Question 16

1. $\frac{70}{120}=\frac{7}{12}$
2. $\frac{45}{50}=\frac{9}{10}$
3. $\frac{45}{120}=\frac{3}{8}$
4. $\frac{45}{70}=\frac{9}{14}$ (correct answer)

Explanation: 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.

Question 17

1. $\frac{25}{50}=0.50$
2. $\frac{25}{90}\approx 0.278$
3. $\frac{40}{90}\approx 0.444$
4. $\frac{25}{40}=0.625$ (correct answer)

Explanation: 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).

Question 18

1. $\frac{42}{72}\approx0.5833$
2. $\frac{80}{160}=0.50$
3. $\frac{42}{80}=0.525$ (correct answer)
4. $\frac{42}{160}=0.2625$

Explanation: 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.

Question 19

1. Yes; $P(A\mid B)=\frac{18}{72}=0.25$ equals $P(A)=\frac{30}{120}=0.25$. (correct answer)
2. Yes; the counts 18 and 54 are close, so the events are independent.
3. No; $P(A\mid B)=\frac{18}{30}=0.60$ does not equal $P(A)=\frac{30}{120}=0.25$.
4. No; $P(A\mid B)=\frac{18}{72}=0.25$ does not equal $P(A)=\frac{72}{120}=0.60$.

Explanation: Two-way tables allow checking for independence by comparing conditional probability to marginal probability. The 'given' for P(A|B) is event B (plays a sport), so the denominator is the sport yes total of 72. The intersection A ∩ B is the cell for job yes and sport yes, which is 18. Compute P(A|B) = 18/72 = 0.25 and P(A) = 30/120 = 0.25; since they equal, the events are independent. A misconception is using the row total 30 as denominator for P(A|B), getting 18/30 = 0.60, or mistakenly using 72/120 as P(A). Strategy: circle the conditioning column (sport yes 72), find the intersection cell (18), form 18/72, then compare to overall P(A) = 30/120. This confirms independence when the probabilities match.

Question 20

1. $\frac{90}{180}=0.50$
2. $\frac{36}{180}=0.20$
3. $\frac{36}{90}=0.40$ (correct answer)
4. $\frac{36}{54}\approx0.67$

Explanation: Conditional probabilities from two-way tables use the given event to set the denominator. For P(Packed lunch | Buys a drink), the 'given' is buys a drink (yes column total 90). The intersection is packed yes and drink yes, cell 36. Thus, 36/90 = 0.40. A distractor might flip to the row total 90 for packed yes, but that's not conditional on drink; another uses grand total 180, giving 36/180 = 0.20, or row for no packed 54/90 ≈ 0.60 but wrong event. Strategy: circle the conditioning column (drink yes 90), find the packed yes cell (36), and compute 36/90. This ensures the fraction reflects the conditioned subgroup.