Conditional Probability and Independence: Formula and Tests
Condition on the correct reference group, distinguish independence from mutual exclusivity, and verify independence with probabilities rather than intuition.
Conditional Probability Independence: direct answer
Conditional probability independence problems change the reference group and ask whether conditioning changes probability. Use P(A|B)=P(A∩B)/P(B), then test independence with a conditional equality or the product criterion.
This page keeps the lesson centered on conditional probability independence. Practice is included only after the method, assumptions, interpretation, and common decision points are explained.
Quick reference: Conditional Probability and Independence: Formula and Tests
| Conditional probability | P(A|B)=P(A∩B)P(B) |
|---|---|
| Independence | P(A|B)=P(A) |
Concept mastery: Conditional Probability and Independence: Formula and Tests
Conditional probability changes the denominator
P(A|B)=P(A∩B)/P(B) when P(B)>0. Once the condition B is imposed, the relevant universe is B rather than the entire sample space. In a two-way table, this means divide by the total in the conditioning row or column, not by the grand total.
Independence can be checked in several equivalent ways
For events with positive probabilities, A and B are independent when P(A|B)=P(A), equivalently P(B|A)=P(B), or P(A∩B)=P(A)P(B). Choose the form that matches the information given. A tiny numerical difference caused by rounding should be interpreted in context rather than treated as exact mathematical evidence.
Tree diagrams encode sequential conditioning
A probability tree places conditional probabilities on branches. Multiply along a path to obtain a joint probability, then add disjoint paths to obtain a broader event. Branch probabilities leaving the same node should sum to 1 because they partition the remaining possibilities at that stage.
Two-way tables make reference groups visible
For categorical data, conditional relative frequencies should be compared across the same response categories while conditioning on different groups. If the conditional distribution of the response is essentially the same in each group, that supports independence. Large systematic differences indicate association.
Mutually exclusive positive-probability events are dependent
If A and B cannot occur together, then P(A|B)=0 whenever P(B)>0. Unless P(A)=0, conditioning on B changes the probability of A, so the events are dependent. This simple argument is stronger than memorizing that exclusivity and independence are “different.”
Association in a two-way table does not establish causation
Conditional distributions can reveal association between categorical variables, but causal interpretation depends on how data were produced. Random assignment can support causal claims about treatments; an observational association can remain confounded even when the conditional percentages differ dramatically.
12 worked conditional probability independence cases
Worked case 1: Table denominator
Scenario. Among 80 commuters, 32 use transit; 20 of those transit users are under 30.
Reasoning. P(under 30 | transit)=20/32, because the condition restricts the denominator to transit users.
Case 1: Table denominator check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 1; do not replace it with a memorized generic sentence.
Worked case 2: Reverse condition
Scenario. Using the same data, 40 commuters are under 30 and 20 use transit.
Reasoning. P(transit | under 30)=20/40, a different denominator and generally a different probability.
Case 2: Reverse condition check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 2; do not replace it with a memorized generic sentence.
Worked case 3: Independence equality
Scenario. P(A)=0.40 and P(A|B)=0.40.
Reasoning. If the probabilities are exact and P(B)>0, this supports independence because conditioning on B does not change A.
Case 3: Independence equality check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 3; do not replace it with a memorized generic sentence.
Worked case 4: Product test
Scenario. P(A)=0.5, P(B)=0.3, P(A∩B)=0.15.
Reasoning. Because 0.5×0.3=0.15, the product criterion for independence holds.
Case 4: Product test check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 4; do not replace it with a memorized generic sentence.
Worked case 5: Dependent events
Scenario. P(A)=0.5 but P(A|B)=0.7.
Reasoning. Conditioning on B changes the probability of A, so the events are dependent.
Case 5: Dependent events check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 5; do not replace it with a memorized generic sentence.
Worked case 6: Exclusive events
Scenario. P(A)>0 and P(B)>0, but P(A∩B)=0.
Reasoning. The events are mutually exclusive and therefore dependent, since independence would require a positive product P(A)P(B).
Case 6: Exclusive events check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 6; do not replace it with a memorized generic sentence.
Worked case 7: Tree path
Scenario. A customer is premium with probability 0.2 and buys given premium with probability 0.6.
Reasoning. The joint probability for the premium-and-buy path is 0.2×0.6=0.12.
Case 7: Tree path check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 7; do not replace it with a memorized generic sentence.
Worked case 8: Conditional distribution
Scenario. Pass rates are 72% in group A and 71% in group B.
Reasoning. The conditional distributions are similar, providing little evidence of a strong association, though sampling variation should be considered.
Case 8: Conditional distribution check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 8; do not replace it with a memorized generic sentence.
Worked case 9: Strong association
Scenario. Pass rates are 82% in group A and 45% in group B.
Reasoning. The large difference in conditional percentages indicates association between group and pass status.
Case 9: Strong association check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 9; do not replace it with a memorized generic sentence.
Worked case 10: Bayes reasoning
Scenario. A positive test can arise from diseased and nondiseased people.
Reasoning. Compute the two joint positive-test paths first; the desired reverse conditional probability uses the diseased-positive path over all positive paths.
Case 10: Bayes reasoning check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 10; do not replace it with a memorized generic sentence.
Worked case 11: Rounding caution
Scenario. A table gives 0.333 and 0.334 after rounding.
Reasoning. Do not declare dependence from a one-thousandth rounding difference without considering the underlying exact counts.
Case 11: Rounding caution check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 11; do not replace it with a memorized generic sentence.
Worked case 12: Causation caution
Scenario. Treatment status and recovery are associated in an observational table.
Reasoning. Conditional association alone does not establish causation because treatment selection may be confounded.
Case 12: Causation caution check: identify the exact statistical target in this scenario, verify the sign, denominator, assignment mechanism, event boundary, or model condition that controls the answer, and end with a conclusion whose scope matches the evidence. This check is specific to case 12; do not replace it with a memorized generic sentence.
Conditional Probability and Independence: Formula and Tests: 36 multiple-choice questions
These questions stay within this page’s topic. Work them after the concept and worked-case sections so practice reinforces the method rather than replacing instruction.
Question 1. Conditional Probability and Independence
In a state park in Midwest consortium during a yearly program evaluation, among Group A, 34 meet and 35 do not meet a criterion; among Group B, 44 meet and 28 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=34(34+35)=0.493. Overall P(meet)=(34+44)(34+35+44+28)=0.553. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 2. Conditional Probability and Independence
In a city recreation department in South Harbor during a summer implementation review, among Group A, 37 meet and 27 do not meet a criterion; among Group B, 41 meet and 48 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=37(37+27)=0.578. Overall P(meet)=(37+41)(37+27+41+48)=0.51. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 3. Conditional Probability and Independence
In a food safety laboratory in Riverbend during a semester-long cohort study, among Group A, 52 meet and 37 do not meet a criterion; among Group B, 30 meet and 35 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=52(52+37)=0.584. Overall P(meet)=(52+30)(52+37+30+35)=0.532. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 4. Conditional Probability and Independence
In a farm cooperative in Coastal Plains during a semester-long cohort study, among Group A, 28 meet and 52 do not meet a criterion; among Group B, 44 meet and 24 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=28(28+52)=0.35. Overall P(meet)=(28+44)(28+52+44+24)=0.486. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 5. Conditional Probability and Independence
In a public high school in Metro East during a yearly program evaluation, among Group A, 72 meet and 15 do not meet a criterion; among Group B, 63 meet and 29 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=72(72+15)=0.828. Overall P(meet)=(72+63)(72+15+63+29)=0.754. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 6. Conditional Probability and Independence
In a state park in Metro East during a weekday operations study, among Group A, 56 meet and 27 do not meet a criterion; among Group B, 22 meet and 40 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=56(56+27)=0.675. Overall P(meet)=(56+22)(56+27+22+40)=0.538. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 7. Conditional Probability and Independence
In a city recreation department in Cedar Grove during a weekday operations study, among Group A, 66 meet and 45 do not meet a criterion; among Group B, 31 meet and 44 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=66(66+45)=0.595. Overall P(meet)=(66+31)(66+45+31+44)=0.522. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 8. Conditional Probability and Independence
In a university advising center in Cedar Grove during a two-month observation window, among Group A, 51 meet and 29 do not meet a criterion; among Group B, 42 meet and 59 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=51(51+29)=0.637. Overall P(meet)=(51+42)(51+29+42+59)=0.514. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 9. Conditional Probability and Independence
In a public health department in Capital Region during a winter readiness review, among Group A, 35 meet and 27 do not meet a criterion; among Group B, 48 meet and 28 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=35(35+27)=0.565. Overall P(meet)=(35+48)(35+27+48+28)=0.601. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 10. Conditional Probability and Independence
In a state park in Central County during a summer implementation review, among Group A, 56 meet and 44 do not meet a criterion; among Group B, 26 meet and 47 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=56(56+44)=0.56. Overall P(meet)=(56+26)(56+44+26+47)=0.474. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 11. Conditional Probability and Independence
In a county library in Pacific Northwest during a community outreach cycle, among Group A, 70 meet and 24 do not meet a criterion; among Group B, 44 meet and 36 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=70(70+24)=0.745. Overall P(meet)=(70+44)(70+24+44+36)=0.655. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 12. Conditional Probability and Independence
In a university advising center in Lakeside district during a pre-exam training cycle, among Group A, 48 meet and 26 do not meet a criterion; among Group B, 43 meet and 47 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: A
P(meet|A)=48(48+26)=0.649. Overall P(meet)=(48+43)(48+26+43+47)=0.555. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 13. Conditional Probability and Independence
In a university advising center in Pine Ridge during a spring 2027 pilot, among Group A, 42 meet and 29 do not meet a criterion; among Group B, 52 meet and 33 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=42(42+29)=0.592. Overall P(meet)=(42+52)(42+29+52+33)=0.603. The probabilities are very close; the table is consistent with approximate independence, though exact independence would require equality.
Question 14. Conditional Probability and Independence
In a solar installer in New England network during a winter readiness review, among Group A, 74 meet and 17 do not meet a criterion; among Group B, 45 meet and 23 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=74(74+17)=0.813. Overall P(meet)=(74+45)(74+17+45+23)=0.748. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 15. Conditional Probability and Independence
In a farm cooperative in Prairie District during a regional benchmarking study, among Group A, 38 meet and 27 do not meet a criterion; among Group B, 23 meet and 38 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=38(38+27)=0.585. Overall P(meet)=(38+23)(38+27+23+38)=0.484. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 16. Conditional Probability and Independence
In a city transit agency in Prairie District during a monthly quality review, among Group A, 69 meet and 16 do not meet a criterion; among Group B, 28 meet and 28 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=69(69+16)=0.812. Overall P(meet)=(69+28)(69+16+28+28)=0.688. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 17. Conditional Probability and Independence
In a solar installer in Westview during a semester-long cohort study, among Group A, 56 meet and 47 do not meet a criterion; among Group B, 65 meet and 45 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=56(56+47)=0.544. Overall P(meet)=(56+65)(56+47+65+45)=0.568. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 18. Conditional Probability and Independence
In a community bank in South Harbor during a monthly quality review, among Group A, 38 meet and 47 do not meet a criterion; among Group B, 48 meet and 58 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=38(38+47)=0.447. Overall P(meet)=(38+48)(38+47+48+58)=0.45. The probabilities are very close; the table is consistent with approximate independence, though exact independence would require equality.
Question 19. Conditional Probability and Independence
In a municipal emergency dispatch center in Metro East during a six-week field trial, among Group A, 52 meet and 19 do not meet a criterion; among Group B, 54 meet and 51 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=52(52+19)=0.732. Overall P(meet)=(52+54)(52+19+54+51)=0.602. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 20. Conditional Probability and Independence
In a city recreation department in South Harbor during a quarterly performance study, among Group A, 65 meet and 40 do not meet a criterion; among Group B, 63 meet and 43 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=65(65+40)=0.619. Overall P(meet)=(65+63)(65+40+63+43)=0.607. The probabilities are very close; the table is consistent with approximate independence, though exact independence would require equality.
Question 21. Conditional Probability and Independence
In a regional manufacturer in North Valley during a baseline measurement week, among Group A, 62 meet and 16 do not meet a criterion; among Group B, 35 meet and 22 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=62(62+16)=0.795. Overall P(meet)=(62+35)(62+16+35+22)=0.719. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 22. Conditional Probability and Independence
In a grocery cooperative in North Valley during a monthly quality review, among Group A, 37 meet and 49 do not meet a criterion; among Group B, 55 meet and 52 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=37(37+49)=0.43. Overall P(meet)=(37+55)(37+49+55+52)=0.477. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 23. Conditional Probability and Independence
In a county election office in Prairie District during a two-month observation window, among Group A, 36 meet and 55 do not meet a criterion; among Group B, 57 meet and 22 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=36(36+55)=0.396. Overall P(meet)=(36+57)(36+55+57+22)=0.547. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 24. Conditional Probability and Independence
In a housing authority in Lakeside district during a pre-exam training cycle, among Group A, 36 meet and 48 do not meet a criterion; among Group B, 52 meet and 40 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=36(36+48)=0.429. Overall P(meet)=(36+52)(36+48+52+40)=0.5. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 25. Conditional Probability and Independence
In a grocery cooperative in Midwest consortium during a weekday operations study, among Group A, 42 meet and 42 do not meet a criterion; among Group B, 30 meet and 38 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=42(42+42)=0.5. Overall P(meet)=(42+30)(42+42+30+38)=0.474. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 26. Conditional Probability and Independence
In a city recreation department in South Harbor during a pre-exam training cycle, among Group A, 51 meet and 29 do not meet a criterion; among Group B, 60 meet and 45 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=51(51+29)=0.637. Overall P(meet)=(51+60)(51+29+60+45)=0.6. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 27. Conditional Probability and Independence
In a wildlife clinic in Desert County during a service-improvement study, among Group A, 31 meet and 17 do not meet a criterion; among Group B, 36 meet and 40 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=31(31+17)=0.646. Overall P(meet)=(31+36)(31+17+36+40)=0.54. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 28. Conditional Probability and Independence
In a county library in Capital Region during a pre-exam training cycle, among Group A, 32 meet and 21 do not meet a criterion; among Group B, 45 meet and 34 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=32(32+21)=0.604. Overall P(meet)=(32+45)(32+21+45+34)=0.583. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 29. Conditional Probability and Independence
In a county election office in Prairie District during a summer implementation review, among Group A, 47 meet and 39 do not meet a criterion; among Group B, 56 meet and 40 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=47(47+39)=0.547. Overall P(meet)=(47+56)(47+39+56+40)=0.566. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 30. Conditional Probability and Independence
In a digital learning platform in Great Lakes during a semester-long cohort study, among Group A, 63 meet and 29 do not meet a criterion; among Group B, 59 meet and 23 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=63(63+29)=0.685. Overall P(meet)=(63+59)(63+29+59+23)=0.701. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 31. Conditional Probability and Independence
In a wildlife clinic in Great Lakes during a community outreach cycle, among Group A, 35 meet and 49 do not meet a criterion; among Group B, 51 meet and 32 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=35(35+49)=0.417. Overall P(meet)=(35+51)(35+49+51+32)=0.515. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 32. Conditional Probability and Independence
In a municipal water office in Pacific Northwest during a fall 2026 audit, among Group A, 37 meet and 17 do not meet a criterion; among Group B, 22 meet and 48 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=37(37+17)=0.685. Overall P(meet)=(37+22)(37+17+22+48)=0.476. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 33. Conditional Probability and Independence
In a school district in Great Lakes during a community outreach cycle, among Group A, 40 meet and 29 do not meet a criterion; among Group B, 35 meet and 25 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: A
P(meet|A)=40(40+29)=0.58. Overall P(meet)=(40+35)(40+29+35+25)=0.581. The probabilities are very close; the table is consistent with approximate independence, though exact independence would require equality.
Question 34. Conditional Probability and Independence
In a farm cooperative in Atlantic Corridor during a community outreach cycle, among Group A, 59 meet and 36 do not meet a criterion; among Group B, 58 meet and 47 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: B
P(meet|A)=59(59+36)=0.621. Overall P(meet)=(59+58)(59+36+58+47)=0.585. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 35. Conditional Probability and Independence
In a university advising center in South Harbor during a quarterly performance study, among Group A, 67 meet and 47 do not meet a criterion; among Group B, 20 meet and 28 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: D
P(meet|A)=67(67+47)=0.588. Overall P(meet)=(67+20)(67+47+20+28)=0.537. The probabilities differ, so meeting the criterion and group membership are not independent.
Question 36. Conditional Probability and Independence
In a housing authority in Coastal Plains during a school-year data collection, among Group A, 31 meet and 21 do not meet a criterion; among Group B, 25 meet and 37 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
Answer: C
P(meet|A)=31(31+21)=0.596. Overall P(meet)=(31+25)(31+21+25+37)=0.491. The probabilities differ, so meeting the criterion and group membership are not independent.
Conditional Probability and Independence: Formula and Tests: 14 free-response questions
For each response, show the statistical reasoning, use the scenario’s language, and state only the conclusion supported by the design or probability model.
FRQ set 1: Conditional Probability and Independence
Scenario. In a regional manufacturer in Westview during a follow-up evaluation period, among Group A, 44 meet and 31 do not meet a criterion; among Group B, 41 meet and 44 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=44(44+31)=0.587. Overall P(meet)=(44+41)(44+31+41+44)=0.531. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 2: Conditional Probability and Independence
Scenario. In a farm cooperative in Atlantic Corridor during a six-week field trial, among Group A, 48 meet and 51 do not meet a criterion; among Group B, 36 meet and 41 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=48(48+51)=0.485. Overall P(meet)=(48+36)(48+51+36+41)=0.477. The probabilities are very close; the table is consistent with approximate independence, though exact independence would require equality.
FRQ set 3: Conditional Probability and Independence
Scenario. In a school district in Coastal Plains during a two-month observation window, among Group A, 40 meet and 29 do not meet a criterion; among Group B, 65 meet and 32 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=40(40+29)=0.58. Overall P(meet)=(40+65)(40+29+65+32)=0.633. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 4: Conditional Probability and Independence
Scenario. In a public high school in Lakeside district during a monthly quality review, among Group A, 68 meet and 50 do not meet a criterion; among Group B, 39 meet and 50 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=68(68+50)=0.576. Overall P(meet)=(68+39)(68+50+39+50)=0.517. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 5: Conditional Probability and Independence
Scenario. In a city recreation department in Cedar Grove during a multiweek validation study, among Group A, 52 meet and 38 do not meet a criterion; among Group B, 58 meet and 33 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=52(52+38)=0.578. Overall P(meet)=(52+58)(52+38+58+33)=0.608. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 6: Conditional Probability and Independence
Scenario. In a municipal water office in Cedar Grove during a two-month observation window, among Group A, 32 meet and 22 do not meet a criterion; among Group B, 29 meet and 25 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=32(32+22)=0.593. Overall P(meet)=(32+29)(32+22+29+25)=0.565. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 7: Conditional Probability and Independence
Scenario. In a public health department in Pine Ridge during a two-month observation window, among Group A, 75 meet and 46 do not meet a criterion; among Group B, 59 meet and 48 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=75(75+46)=0.62. Overall P(meet)=(75+59)(75+46+59+48)=0.588. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 8: Conditional Probability and Independence
Scenario. In a city recreation department in Pine Ridge during a six-week field trial, among Group A, 51 meet and 27 do not meet a criterion; among Group B, 55 meet and 38 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=51(51+27)=0.654. Overall P(meet)=(51+55)(51+27+55+38)=0.62. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 9: Conditional Probability and Independence
Scenario. In a wildlife clinic in North Valley during a service-improvement study, among Group A, 62 meet and 53 do not meet a criterion; among Group B, 44 meet and 37 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=62(62+53)=0.539. Overall P(meet)=(62+44)(62+53+44+37)=0.541. The probabilities are very close; the table is consistent with approximate independence, though exact independence would require equality.
FRQ set 10: Conditional Probability and Independence
Scenario. In a solar installer in Pacific Northwest during a summer implementation review, among Group A, 71 meet and 51 do not meet a criterion; among Group B, 41 meet and 49 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=71(71+51)=0.582. Overall P(meet)=(71+41)(71+51+41+49)=0.528. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 11: Conditional Probability and Independence
Scenario. In a solar installer in Coastal Plains during a follow-up evaluation period, among Group A, 67 meet and 27 do not meet a criterion; among Group B, 51 meet and 43 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=67(67+27)=0.713. Overall P(meet)=(67+51)(67+27+51+43)=0.628. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 12: Conditional Probability and Independence
Scenario. In a digital learning platform in Mountain Region during a six-week field trial, among Group A, 68 meet and 38 do not meet a criterion; among Group B, 30 meet and 21 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=68(68+38)=0.642. Overall P(meet)=(68+30)(68+38+30+21)=0.624. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 13: Conditional Probability and Independence
Scenario. In a housing authority in Pacific Northwest during a monthly quality review, among Group A, 37 meet and 55 do not meet a criterion; among Group B, 39 meet and 36 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=37(37+55)=0.402. Overall P(meet)=(37+39)(37+55+39+36)=0.455. The probabilities differ, so meeting the criterion and group membership are not independent.
FRQ set 14: Conditional Probability and Independence
Scenario. In a recycling program in Pine Ridge during a weekday operations study, among Group A, 67 meet and 16 do not meet a criterion; among Group B, 59 meet and 55 do not. Find P(meet|A) and compare it with P(meet) to assess independence.
- Define the event or random variable and state the requested probability or long-run quantity.
- Verify the relevant model conditions or probability-distribution requirements.
- Show the probability, expected-value, or distribution calculation with labeled terms.
- Interpret the result as a probability or long-run behavior and distinguish it from a guarantee for one trial.
Model response
P(meet|A)=67(67+16)=0.807. Overall P(meet)=(67+59)(67+16+59+55)=0.64. The probabilities differ, so meeting the criterion and group membership are not independent.
Continue with a connected AP Statistics skill
Use the next page to connect this method to a neighboring concept rather than repeating the same calculation pattern.
