Using Two-Way Tables for Probability
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Statistics › Using Two-Way Tables for Probability
A random sample of 150 students recorded whether they prefer to study in the morning or evening and whether they usually study alone or in a group. Using the two-way table, what is $P(\text{Morning}\mid\text{Group})$? Give your answer as a fraction.
Two-way table (counts):
- Group & Morning: 18
- Group & Evening: 42
- Alone & Morning: 45
- Alone & Evening: 45
- Row totals: Group = 60, Alone = 90
- Column totals: Morning = 63, Evening = 87
- Grand total: 150
$\frac{18}{150}$
$\frac{18}{60}$
$\frac{18}{63}$
$\frac{63}{60}$
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 $\frac{18}{60}$.
A random sample of 180 students was asked whether they have a part-time job (Yes/No) and whether they play a school sport (Yes/No). Using the two-way table, estimate the probability that a randomly selected student plays a school sport given that the student has a part-time job. Give your answer as a percent.
Two-way table (counts):
- Job Yes & Sport Yes: 27
- Job Yes & Sport No: 63
- Job No & Sport Yes: 54
- Job No & Sport No: 36
- Row totals: Job Yes = 90, Job No = 90
- Column totals: Sport Yes = 81, Sport No = 99
- Grand total: 180
15%
30%
45%
50%
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%.
A random sample of 96 students was asked whether they ride the bus or get a car ride to school, and whether they usually arrive on time or late. Using the two-way table, estimate the probability that a randomly selected student arrives on time given that the student rides the bus. Give your answer as a fraction.
Two-way table (counts):
- Bus & On time: 30
- Bus & Late: 18
- Car & On time: 40
- Car & Late: 8
- Row totals: Bus = 48, Car = 48
- Column totals: On time = 70, Late = 26
- Grand total: 96
$\frac{30}{70}$
$\frac{30}{96}$
$\frac{48}{70}$
$\frac{30}{48}$
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.
A random sample of 100 students reported whether they usually do homework at home or at school, and whether they use paper notes or digital notes. Using the two-way table, are the events A = “student uses digital notes” and B = “student does homework at school” independent? Use the table to justify by comparing $P(A\mid B)$ to $P(A)$.
Two-way table (counts):
- At school & Digital: 18
- At school & Paper: 12
- At home & Digital: 42
- At home & Paper: 28
- Row totals: At school = 30, At home = 70
- Column totals: Digital = 60, Paper = 40
- Grand total: 100
No; because 70 students do homework at home, the events cannot be independent.
Yes; $\frac{18}{100}$ equals $\frac{30}{100}$, so they are independent.
Yes; $P(\text{Digital}\mid\text{At school})=\frac{18}{30}$ equals $P(\text{Digital})=\frac{60}{100}$.
No; $P(\text{Digital}\mid\text{At school})=\frac{18}{30}$ does not equal $P(\text{Digital})=\frac{60}{100}$.
Explanation
Independence in two-way tables is assessed by checking if a conditional probability equals the marginal probability. For P(Digital | At school), the 'at school' total is 30 as denominator, and intersection with digital is 18, so 18/30 = 0.6. The marginal P(Digital) is 60/100 = 0.6, which matches, indicating independence. A distractor might compare joint probabilities like 18/100 to 30/100, but that doesn't test independence properly. Circle the 'at school' total 30, compute 18/30, and compare to 60/100; since equal, the events are independent.
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
20%
30%
40%
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%.
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
0.36
0.50
0.71
0.83
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.
A random sample of 120 students at a high school was asked which device they use most for schoolwork (Laptop or Tablet) and whether they participate in an after-school club (Yes or No). Using the two-way table, estimate the probability that a randomly selected student uses a laptop given that the student is in a club. Give your answer as a decimal.
Two-way table (counts):
- Club Yes & Laptop: 36
- Club Yes & Tablet: 24
- Club No & Laptop: 30
- Club No & Tablet: 30
- Row totals: Club Yes = 60, Club No = 60
- Column totals: Laptop = 66, Tablet = 54
- Grand total: 120
0.30
0.50
0.55
0.60
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 for that group, which is 60 students in clubs. The numerator is the cell where the student is in a club and uses a laptop, which is 36. Thus, the conditional probability is 36/60 = 0.60. A common misconception is using the grand total as the denominator, which would give 36/120 = 0.30, but that's the joint probability, not conditional. To avoid errors, circle the 'club yes' row total of 60, then divide the intersection cell of 36 by that total to form the fraction.
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
$\frac{65}{80}$
$\frac{35}{160}$
$\frac{35}{65}$
$\frac{35}{80}$
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.
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
Yes; $P(\text{CS Yes}\mid\text{Team})=\frac{28}{70}$ equals $P(\text{CS Yes})=\frac{60}{150}$.
No; $P(\text{CS Yes}\mid\text{Team})=\frac{28}{70}$ does not equal $P(\text{CS Yes})=\frac{60}{150}$.
No; because $\frac{28}{150}$ is not equal to $\frac{32}{150}$.
Yes; because 70 and 80 are close, the events are independent.
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.
A random sample of 110 students recorded whether they attend after-school tutoring and whether they passed the most recent math quiz. Using the two-way table, estimate the probability that a randomly selected student passed given that the student attends tutoring. Give your answer as a percent.
Two-way table (counts):
- Tutoring & Passed: 32
- Tutoring & Did not pass: 18
- No tutoring & Passed: 45
- No tutoring & Did not pass: 15 Row totals: Tutoring 50, No tutoring 60 Column totals: Passed 77, Did not pass 33 Grand total: 110
$41.6%$
$70%$
$64%$
$29%$
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.