Finite Conditional Probability Foundations
This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
3/7
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
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 explanation
0/1
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
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 explanation
11/12
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
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
1/2
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
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
1/5
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
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 explanation
0/1
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
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 explanation
21/23
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
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
1/3
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
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 explanation
0/1
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
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 explanation
26/27
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
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
3/7
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
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 explanation
0/1
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
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 explanation
1/1
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
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 explanation
0/1
|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
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 explanation
1/1
|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
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 explanation
0/1
|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
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 explanation
3/4
|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
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 explanation
1/1
|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
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 explanation
1/2
|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
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 explanation
0/1
|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
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 explanation
1/1
|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
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 explanation
0/1
|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
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 explanation
3/4
|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
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 explanation
1/1
|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
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 explanation
1/2
|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
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 explanation
0/1
|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
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 explanation
1/1
|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
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 explanation
0/1
|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
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 explanation
3/4
|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
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 explanation
1/1
|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
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 explanation
1/2
|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
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 explanation
0/1
|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
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 explanation
1/1
|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
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 explanation
0/1
|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
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 explanation
3/4
|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
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 explanation
1/1
|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
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 explanation
1/2
|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
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 explanation
0/1
|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
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 explanation
No
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
Yes
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
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 explanation
No
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
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 explanation
No
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
Conditional probability in finite models · qf-conditional-direction-comparison · candidate
Canonical question JSONL · English JSONL · Validation report