# Finite Conditional Probability Foundations

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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

batch-20261003-advance-20261004-r07-conditional-probability-foundations · 2026-10-03 · 100 questions

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.



## 1. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(4), 1/(4), 2/(4), 1/(4)], summing exactly to one. A=∅, B={-12, -11, -10, -9}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 4. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0001/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 2. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-11, -10, -9, -8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(8), 2/(8), 3/(8), 2/(8)], summing exactly to one. A={-11, -10, -9, -8}, B={-11, -10, -9, -8}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 8 and the B mass numerator is 8. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0002/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 3. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-10, -9, -8, -7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 3/(12), 4/(12), 3/(12)], summing exactly to one. A={-10, -9}, B={-9, -8}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation3/7

The intersection mass numerator is 3 and the B mass numerator is 7. Their common positive denominator cancels, giving P(A|B)=3/7; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0003/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 4. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 4/(13), 5/(13), 4/(13)], summing exactly to one. A={-9}, B={-7, -6}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 9. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0004/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 5. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-8, -7, -6, -5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(13), 1/(13), 6/(13), 5/(13)], summing exactly to one. A={-6, -5}, B={-8, -6, -5}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation11/12

The intersection mass numerator is 11 and the B mass numerator is 12. Their common positive denominator cancels, giving P(A|B)=11/12; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0005/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 6. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-7, -6, -5, -4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 2/(12), 2/(12), 6/(12)], summing exactly to one. A=∅, B={-7, -6, -5, -4}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 12. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0006/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 7. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 3/(13), 3/(13), 7/(13)], summing exactly to one. A={-6, -5, -4, -3}, B={-6, -5, -4, -3}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 13 and the B mass numerator is 13. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0007/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 8. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-5, -4, -3, -2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 4/(17), 4/(17), 8/(17)], summing exactly to one. A={-5, -4}, B={-4, -3}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/2

The intersection mass numerator is 4 and the B mass numerator is 8. Their common positive denominator cancels, giving P(A|B)=1/2; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0008/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 9. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-4, -3, -2, -1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(17), 1/(17), 5/(17), 9/(17)], summing exactly to one. A={-4}, B={-2, -1}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 14. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0009/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 10. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 2/(18), 6/(18), 10/(18)], summing exactly to one. A={-1, 0}, B={-3, -1, 0}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 16 and the B mass numerator is 16. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0010/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 11. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-2, -1, 0, 1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 3/(17), 2/(17), 11/(17)], summing exactly to one. A=∅, B={-2, -1, 0, 1}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 17. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0011/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 12. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={-1, 0, 1, 2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(21), 4/(21), 3/(21), 12/(21)], summing exactly to one. A={-1, 0, 1, 2}, B={-1, 0, 1, 2}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 21 and the B mass numerator is 21. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0012/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 13. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 1/(18), 4/(18), 13/(18)], summing exactly to one. A={0, 1}, B={1, 2}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/5

The intersection mass numerator is 1 and the B mass numerator is 5. Their common positive denominator cancels, giving P(A|B)=1/5; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0013/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 14. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={1, 2, 3, 4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 2/(22), 5/(22), 14/(22)], summing exactly to one. A={1}, B={3, 4}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 19. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0014/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 15. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={2, 3, 4, 5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 3/(26), 6/(26), 15/(26)], summing exactly to one. A={4, 5}, B={2, 4, 5}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation21/23

The intersection mass numerator is 21 and the B mass numerator is 23. Their common positive denominator cancels, giving P(A|B)=21/23; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0015/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 16. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(22), 4/(22), 2/(22), 16/(22)], summing exactly to one. A=∅, B={3, 4, 5, 6}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 22. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0016/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 17. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={4, 5, 6, 7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 1/(22), 3/(22), 17/(22)], summing exactly to one. A={4, 5, 6, 7}, B={4, 5, 6, 7}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 22 and the B mass numerator is 22. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0017/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 18. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={5, 6, 7, 8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 2/(26), 4/(26), 18/(26)], summing exactly to one. A={5, 6}, B={6, 7}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/3

The intersection mass numerator is 2 and the B mass numerator is 6. Their common positive denominator cancels, giving P(A|B)=1/3; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0018/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 19. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 3/(27), 5/(27), 19/(27)], summing exactly to one. A={6}, B={8, 9}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 24. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0019/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 20. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={7, 8, 9, 10}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 4/(31), 6/(31), 20/(31)], summing exactly to one. A={9, 10}, B={7, 9, 10}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation26/27

The intersection mass numerator is 26 and the B mass numerator is 27. Their common positive denominator cancels, giving P(A|B)=26/27; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0020/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 21. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={8, 9, 10, 11}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 1/(26), 2/(26), 21/(26)], summing exactly to one. A=∅, B={8, 9, 10, 11}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 26. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0021/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 22. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 2/(27), 3/(27), 22/(27)], summing exactly to one. A={9, 10, 11, 12}, B={9, 10, 11, 12}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 27 and the B mass numerator is 27. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0022/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 23. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={10, 11, 12, 13}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 3/(31), 4/(31), 23/(31)], summing exactly to one. A={10, 11}, B={11, 12}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation3/7

The intersection mass numerator is 3 and the B mass numerator is 7. Their common positive denominator cancels, giving P(A|B)=3/7; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0023/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 24. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={11, 12, 13, 14}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(35), 4/(35), 5/(35), 24/(35)], summing exactly to one. A={11}, B={13, 14}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation0/1

The intersection mass numerator is 0 and the B mass numerator is 29. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0024/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 25. Conditional Probability in a Finite Weighted Model

The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(32), 1/(32), 6/(32), 25/(32)], summing exactly to one. A={14, 15}, B={12, 14, 15}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.

View answer and explanation1/1

The intersection mass numerator is 31 and the B mass numerator is 31. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0025/)

Conditional probability in finite models · qf-conditional-weighted-probability · candidate





## 26. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={-12, -11, -10, -9} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0026/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 27. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-11, -10, -9, -8, -7}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-11, -10, -9, -8, -7} and B={-11, -10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0027/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 28. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-10, -9, -8, -7, -6, -5}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-10} and B={-6, -5} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0028/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 29. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-9, -8, -7} and B={-9, -8, -7, -6} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation3/4

|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0029/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 30. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-8, -7, -6, -5, -4}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-8, -7} and B={-8, -7} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0030/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 31. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-7, -6, -5, -4, -3, -2}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-7, -5, -3} and B={-3, -2} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/2

|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0031/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 32. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={-6, -5, -4, -3} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0032/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 33. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-5, -4, -3, -2, -1}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-5, -4, -3, -2, -1} and B={-5, -4} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0033/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 34. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-4, -3, -2, -1, 0, 1}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-4} and B={0, 1} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0034/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 35. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-3, -2, -1} and B={-3, -2, -1, 0} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation3/4

|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0035/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 36. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-2, -1, 0, 1, 2}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-2, -1} and B={-2, -1} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0036/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 37. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={-1, 0, 1, 2, 3, 4}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-1, 1, 3} and B={3, 4} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/2

|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0037/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 38. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={0, 1, 2, 3} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0038/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 39. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={1, 2, 3, 4, 5}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={1, 2, 3, 4, 5} and B={1, 2} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0039/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 40. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={2, 3, 4, 5, 6, 7}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={2} and B={6, 7} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0040/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 41. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={3, 4, 5} and B={3, 4, 5, 6} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation3/4

|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0041/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 42. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={4, 5, 6, 7, 8}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={4, 5} and B={4, 5} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0042/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 43. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={5, 6, 7, 8, 9, 10}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={5, 7, 9} and B={9, 10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/2

|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0043/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 44. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={6, 7, 8, 9} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0044/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 45. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={7, 8, 9, 10, 11}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={7, 8, 9, 10, 11} and B={7, 8} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0045/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 46. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={8, 9, 10, 11, 12, 13}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={8} and B={12, 13} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0046/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 47. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={9, 10, 11} and B={9, 10, 11, 12} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation3/4

|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0047/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 48. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={10, 11, 12, 13, 14}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={10, 11} and B={10, 11} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/1

|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0048/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 49. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={11, 12, 13, 14, 15, 16}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={11, 13, 15} and B={15, 16} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation1/2

|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0049/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 50. Conditional Probability with Explicit Equiprobability

The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={12, 13, 14, 15} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

View answer and explanation0/1

|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0050/)

Conditional probability in finite models · qf-conditional-uniform-probability · candidate





## 51. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(4), 1/(4), 2/(4), 1/(4)], summing exactly to one. Does B={-12} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 4. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0051/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 52. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-11, -10, -9, -8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(8), 2/(8), 3/(8), 2/(8)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 8. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0052/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 53. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-10, -9, -8, -7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 3/(12), 4/(12), 3/(12)], summing exactly to one. Does B={-9} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 3, with positive common denominator 12. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0053/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 54. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 4/(13), 5/(13), 4/(13)], summing exactly to one. Does B={-9, -8, -7, -6} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 13, with positive common denominator 13. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0054/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 55. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-8, -7, -6, -5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(13), 1/(13), 6/(13), 5/(13)], summing exactly to one. Does B={-8} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 1, with positive common denominator 13. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0055/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 56. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-7, -6, -5, -4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 2/(12), 2/(12), 6/(12)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 12. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0056/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 57. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 3/(13), 3/(13), 7/(13)], summing exactly to one. Does B={-5} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 3, with positive common denominator 13. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0057/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 58. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-5, -4, -3, -2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 4/(17), 4/(17), 8/(17)], summing exactly to one. Does B={-5, -4, -3, -2} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 17, with positive common denominator 17. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0058/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 59. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-4, -3, -2, -1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(17), 1/(17), 5/(17), 9/(17)], summing exactly to one. Does B={-4} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 2, with positive common denominator 17. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0059/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 60. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 2/(18), 6/(18), 10/(18)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 18. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0060/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 61. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-2, -1, 0, 1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 3/(17), 2/(17), 11/(17)], summing exactly to one. Does B={-1} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 3, with positive common denominator 17. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0061/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 62. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={-1, 0, 1, 2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(21), 4/(21), 3/(21), 12/(21)], summing exactly to one. Does B={-1, 0, 1, 2} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 21, with positive common denominator 21. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0062/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 63. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 1/(18), 4/(18), 13/(18)], summing exactly to one. Does B={0} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 18. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0063/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 64. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={1, 2, 3, 4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 2/(22), 5/(22), 14/(22)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 22. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0064/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 65. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={2, 3, 4, 5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 3/(26), 6/(26), 15/(26)], summing exactly to one. Does B={3} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 3, with positive common denominator 26. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0065/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 66. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(22), 4/(22), 2/(22), 16/(22)], summing exactly to one. Does B={3, 4, 5, 6} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 22, with positive common denominator 22. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0066/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 67. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={4, 5, 6, 7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 1/(22), 3/(22), 17/(22)], summing exactly to one. Does B={4} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 1, with positive common denominator 22. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0067/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 68. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={5, 6, 7, 8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 2/(26), 4/(26), 18/(26)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 26. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0068/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 69. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 3/(27), 5/(27), 19/(27)], summing exactly to one. Does B={7} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 3, with positive common denominator 27. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0069/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 70. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={7, 8, 9, 10}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 4/(31), 6/(31), 20/(31)], summing exactly to one. Does B={7, 8, 9, 10} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 31, with positive common denominator 31. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0070/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 71. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={8, 9, 10, 11}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 1/(26), 2/(26), 21/(26)], summing exactly to one. Does B={8} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 2, with positive common denominator 26. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0071/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 72. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 2/(27), 3/(27), 22/(27)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 27. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0072/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 73. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={10, 11, 12, 13}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 3/(31), 4/(31), 23/(31)], summing exactly to one. Does B={11} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 3, with positive common denominator 31. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0073/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 74. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={11, 12, 13, 14}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(35), 4/(35), 5/(35), 24/(35)], summing exactly to one. Does B={11, 12, 13, 14} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationYes

The B mass numerator is 35, with positive common denominator 35. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0074/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 75. Positive-Probability Eligibility of a Conditioning Event

The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(32), 1/(32), 6/(32), 25/(32)], summing exactly to one. Does B={12} meet the positive-probability requirement for defining P(A|B)?

View answer and explanationNo

The B mass numerator is 0, with positive common denominator 32. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0075/)

Conditional probability in finite models · qf-conditional-event-validity · candidate





## 76. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(4), 1/(4), 2/(4), 1/(4)], summing exactly to one. A={-11}, B={-10}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0076/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 77. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-11, -10, -9, -8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(8), 2/(8), 3/(8), 2/(8)], summing exactly to one. A={-10, -9}, B={-9, -8}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=3/5 and P(B|A)=3/5, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0077/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 78. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-10, -9, -8, -7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 3/(12), 4/(12), 3/(12)], summing exactly to one. A={-9, -8}, B={-9, -8}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0078/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 79. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 4/(13), 5/(13), 4/(13)], summing exactly to one. A={-9, -8, -7, -6}, B={-8, -7}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=1/1 and P(B|A)=9/13, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0079/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 80. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-8, -7, -6, -5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(13), 1/(13), 6/(13), 5/(13)], summing exactly to one. A={-7, -5}, B={-8, -7, -6, -5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=6/13 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0080/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 81. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-7, -6, -5, -4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 2/(12), 2/(12), 6/(12)], summing exactly to one. A={-6}, B={-5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0081/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 82. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 3/(13), 3/(13), 7/(13)], summing exactly to one. A={-5, -4}, B={-4, -3}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=3/10 and P(B|A)=1/2, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0082/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 83. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-5, -4, -3, -2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 4/(17), 4/(17), 8/(17)], summing exactly to one. A={-4, -3}, B={-4, -3}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0083/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 84. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-4, -3, -2, -1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(17), 1/(17), 5/(17), 9/(17)], summing exactly to one. A={-4, -3, -2, -1}, B={-3, -2}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=1/1 and P(B|A)=6/17, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0084/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 85. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 2/(18), 6/(18), 10/(18)], summing exactly to one. A={-2, 0}, B={-3, -2, -1, 0}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=2/3 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0085/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 86. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-2, -1, 0, 1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 3/(17), 2/(17), 11/(17)], summing exactly to one. A={-1}, B={0}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0086/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 87. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={-1, 0, 1, 2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(21), 4/(21), 3/(21), 12/(21)], summing exactly to one. A={0, 1}, B={1, 2}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=1/5 and P(B|A)=3/7, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0087/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 88. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 1/(18), 4/(18), 13/(18)], summing exactly to one. A={1, 2}, B={1, 2}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0088/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 89. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={1, 2, 3, 4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 2/(22), 5/(22), 14/(22)], summing exactly to one. A={1, 2, 3, 4}, B={2, 3}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=1/1 and P(B|A)=7/22, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0089/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 90. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={2, 3, 4, 5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 3/(26), 6/(26), 15/(26)], summing exactly to one. A={3, 5}, B={2, 3, 4, 5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=9/13 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0090/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 91. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(22), 4/(22), 2/(22), 16/(22)], summing exactly to one. A={4}, B={5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0091/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 92. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={4, 5, 6, 7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 1/(22), 3/(22), 17/(22)], summing exactly to one. A={5, 6}, B={6, 7}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=3/20 and P(B|A)=3/4, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0092/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 93. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={5, 6, 7, 8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 2/(26), 4/(26), 18/(26)], summing exactly to one. A={6, 7}, B={6, 7}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0093/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 94. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 3/(27), 5/(27), 19/(27)], summing exactly to one. A={6, 7, 8, 9}, B={7, 8}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=1/1 and P(B|A)=8/27, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0094/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 95. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={7, 8, 9, 10}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 4/(31), 6/(31), 20/(31)], summing exactly to one. A={8, 10}, B={7, 8, 9, 10}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=24/31 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0095/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 96. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={8, 9, 10, 11}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 1/(26), 2/(26), 21/(26)], summing exactly to one. A={9}, B={10}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0096/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 97. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 2/(27), 3/(27), 22/(27)], summing exactly to one. A={10, 11}, B={11, 12}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=3/25 and P(B|A)=3/5, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0097/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 98. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={10, 11, 12, 13}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 3/(31), 4/(31), 23/(31)], summing exactly to one. A={11, 12}, B={11, 12}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationYes

P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0098/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 99. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={11, 12, 13, 14}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(35), 4/(35), 5/(35), 24/(35)], summing exactly to one. A={11, 12, 13, 14}, B={12, 13}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=1/1 and P(B|A)=9/35, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0099/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate





## 100. Comparing Two Directions of Conditional Probability

The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(32), 1/(32), 6/(32), 25/(32)], summing exactly to one. A={13, 15}, B={12, 13, 14, 15}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?

View answer and explanationNo

P(A|B)=13/16 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence.



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261003-advance-20261004-r07-conditional-probability-0100/)

Conditional probability in finite models · qf-conditional-direction-comparison · candidate



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