Finite Conditional Laws and Candidate Defects
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batch-20261008-advance-20261010-r04-conditional-probability-foundations · 2026-10-08 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Intersection and Conditioning-Event Probabilities
S={0,…,2}, F=P(S), masses=[0, 1, 2], D=3, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[], B=[1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(0/1, 1/3)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (0/1, 1/3).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
2. Intersection and Conditioning-Event Probabilities
S={0,…,2}, F=P(S), masses=[1, 0, 0], D=1, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(1/1, 1/1)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (1/1, 1/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
3. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[2, 1, 2, 0], D=5, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(2/5, 3/5)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (2/5, 3/5).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
4. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[3, 1, 2, 0], D=6, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(1/2, 2/3)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (1/2, 2/3).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
5. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[4, 1, 2, 0], D=7, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(4/7, 5/7)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (4/7, 5/7).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
6. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[5, 1, 2, 0], D=8, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(5/8, 3/4)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (5/8, 3/4).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
7. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[6, 1, 2, 0], D=9, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(2/3, 7/9)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (2/3, 7/9).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
8. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[7, 1, 2, 0], D=10, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(7/10, 4/5)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (7/10, 4/5).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
9. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[8, 1, 2, 0], D=11, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(8/11, 9/11)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (8/11, 9/11).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
10. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[9, 1, 2, 0], D=12, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(3/4, 5/6)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (3/4, 5/6).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
11. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[10, 1, 2, 0], D=13, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(10/13, 11/13)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (10/13, 11/13).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
12. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[11, 1, 2, 0], D=14, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(11/14, 6/7)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (11/14, 6/7).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
13. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[12, 1, 2, 0], D=15, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(4/5, 13/15)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (4/5, 13/15).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
14. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[13, 1, 2, 0], D=16, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(13/16, 7/8)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (13/16, 7/8).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
15. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[14, 1, 2, 0], D=17, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(14/17, 15/17)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (14/17, 15/17).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
16. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[15, 1, 2, 0], D=18, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(5/6, 8/9)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (5/6, 8/9).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
17. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[16, 1, 2, 0], D=19, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(16/19, 17/19)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (16/19, 17/19).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
18. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[17, 1, 2, 0], D=20, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(17/20, 9/10)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (17/20, 9/10).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
19. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[18, 1, 2, 0], D=21, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(6/7, 19/21)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (6/7, 19/21).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
20. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[19, 1, 2, 0], D=22, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(19/22, 10/11)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (19/22, 10/11).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
21. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[20, 1, 2, 0], D=23, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(20/23, 21/23)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (20/23, 21/23).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
22. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[21, 1, 2, 0], D=24, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(7/8, 11/12)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (7/8, 11/12).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
23. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[22, 1, 2, 0], D=25, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(22/25, 23/25)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (22/25, 23/25).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
24. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[23, 1, 2, 0], D=26, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(23/26, 12/13)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (23/26, 12/13).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
25. Intersection and Conditioning-Event Probabilities
S={0,…,3}, F=P(S), masses=[24, 1, 2, 0], D=27, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Give (P(A∩B), P(B)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(8/9, 25/27)
Sum each event mass and divide separately by D; this question does not yet divide the two probabilities, and nonempty B alone would not imply positive probability. The exact result, retaining the requested order, is (8/9, 25/27).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-masses-pair · candidate
26. Both Conditional-Probability Directions
S={0,…,1}, F=P(S), masses=[1, 2], D=3, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0], B=[1], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(0/1, 0/1)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (0/1, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
27. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[1, 0, 2], D=3, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0], B=[0, 1], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(1/1, 1/1)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (1/1, 1/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
28. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[2, 1, 2], D=5, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(1/2, 2/3)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (1/2, 2/3).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
29. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[3, 1, 2], D=6, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(3/5, 3/4)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (3/5, 3/4).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
30. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[4, 1, 2], D=7, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(2/3, 4/5)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (2/3, 4/5).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
31. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[5, 1, 2], D=8, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(5/7, 5/6)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (5/7, 5/6).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
32. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[6, 1, 2], D=9, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(3/4, 6/7)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (3/4, 6/7).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
33. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[7, 1, 2], D=10, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(7/9, 7/8)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (7/9, 7/8).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
34. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[8, 1, 2], D=11, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(4/5, 8/9)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (4/5, 8/9).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
35. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[9, 1, 2], D=12, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(9/11, 9/10)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (9/11, 9/10).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
36. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[10, 1, 2], D=13, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(5/6, 10/11)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (5/6, 10/11).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
37. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[11, 1, 2], D=14, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(11/13, 11/12)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (11/13, 11/12).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
38. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[12, 1, 2], D=15, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(6/7, 12/13)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (6/7, 12/13).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
39. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[13, 1, 2], D=16, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(13/15, 13/14)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (13/15, 13/14).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
40. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[14, 1, 2], D=17, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(7/8, 14/15)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (7/8, 14/15).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
41. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[15, 1, 2], D=18, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(15/17, 15/16)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (15/17, 15/16).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
42. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[16, 1, 2], D=19, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(8/9, 16/17)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (8/9, 16/17).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
43. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[17, 1, 2], D=20, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(17/19, 17/18)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (17/19, 17/18).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
44. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[18, 1, 2], D=21, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(9/10, 18/19)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (9/10, 18/19).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
45. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[19, 1, 2], D=22, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(19/21, 19/20)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (19/21, 19/20).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
46. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[20, 1, 2], D=23, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(10/11, 20/21)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (10/11, 20/21).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
47. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[21, 1, 2], D=24, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(21/23, 21/22)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (21/23, 21/22).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
48. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[22, 1, 2], D=25, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(11/12, 22/23)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (11/12, 22/23).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
49. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[23, 1, 2], D=26, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(23/25, 23/24)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (23/25, 23/24).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
50. Both Conditional-Probability Directions
S={0,…,2}, F=P(S), masses=[24, 1, 2], D=27, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 1], B=[0, 2], P(A)>0 and P(B)>0. Give (P(A|B), P(B|A)). Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
(12/13, 24/25)
Divide the shared intersection mass by the B mass and the A mass respectively; the two directions generally differ, and independence is not assumed. The exact result, retaining the requested order, is (12/13, 24/25).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-directions-pair · candidate
51. Complete Pointwise Conditional Law
S={0,…,2}, F=P(S), masses=[0, 1, 2], D=3, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[1], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[0/1, 1/1, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [0/1, 1/1, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
52. Complete Pointwise Conditional Law
S={0,…,0}, F=P(S), masses=[1], D=1, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[1/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [1/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
53. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[2, 1, 2, 3], D=8, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[2/5, 1/5, 2/5, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [2/5, 1/5, 2/5, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
54. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[3, 1, 2, 3], D=9, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[1/2, 1/6, 1/3, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [1/2, 1/6, 1/3, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
55. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[4, 1, 2, 3], D=10, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[4/7, 1/7, 2/7, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [4/7, 1/7, 2/7, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
56. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[5, 1, 2, 3], D=11, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[5/8, 1/8, 1/4, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [5/8, 1/8, 1/4, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
57. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[6, 1, 2, 3], D=12, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[2/3, 1/9, 2/9, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [2/3, 1/9, 2/9, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
58. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[7, 1, 2, 3], D=13, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[7/10, 1/10, 1/5, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [7/10, 1/10, 1/5, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
59. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[8, 1, 2, 3], D=14, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[8/11, 1/11, 2/11, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [8/11, 1/11, 2/11, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
60. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[9, 1, 2, 3], D=15, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[3/4, 1/12, 1/6, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [3/4, 1/12, 1/6, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
61. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[10, 1, 2, 3], D=16, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[10/13, 1/13, 2/13, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [10/13, 1/13, 2/13, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
62. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[11, 1, 2, 3], D=17, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[11/14, 1/14, 1/7, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [11/14, 1/14, 1/7, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
63. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[12, 1, 2, 3], D=18, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[4/5, 1/15, 2/15, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [4/5, 1/15, 2/15, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
64. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[13, 1, 2, 3], D=19, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[13/16, 1/16, 1/8, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [13/16, 1/16, 1/8, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
65. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[14, 1, 2, 3], D=20, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[14/17, 1/17, 2/17, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [14/17, 1/17, 2/17, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
66. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[15, 1, 2, 3], D=21, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[5/6, 1/18, 1/9, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [5/6, 1/18, 1/9, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
67. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[16, 1, 2, 3], D=22, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[16/19, 1/19, 2/19, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [16/19, 1/19, 2/19, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
68. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[17, 1, 2, 3], D=23, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[17/20, 1/20, 1/10, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [17/20, 1/20, 1/10, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
69. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[18, 1, 2, 3], D=24, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[6/7, 1/21, 2/21, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [6/7, 1/21, 2/21, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
70. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[19, 1, 2, 3], D=25, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[19/22, 1/22, 1/11, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [19/22, 1/22, 1/11, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
71. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[20, 1, 2, 3], D=26, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[20/23, 1/23, 2/23, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [20/23, 1/23, 2/23, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
72. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[21, 1, 2, 3], D=27, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[7/8, 1/24, 1/12, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [7/8, 1/24, 1/12, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
73. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[22, 1, 2, 3], D=28, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[22/25, 1/25, 2/25, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [22/25, 1/25, 2/25, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
74. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[23, 1, 2, 3], D=29, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[23/26, 1/26, 1/13, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [23/26, 1/26, 1/13, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
75. Complete Pointwise Conditional Law
S={0,…,3}, F=P(S), masses=[24, 1, 2, 3], D=30, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. B=[0, 1, 2], P(B)>0. In index order from 0, list the conditional probability of every singleton given B, including zero terms outside B. Use exact reduced fractions with positive denominators, retaining /1 for integers.
View answer and explanation
[8/9, 1/27, 2/27, 0/1]
For a point in B divide its mass by the total B mass; outside B use 0/1. The complete vector sums to one. The exact result, retaining the requested order, is [8/9, 1/27, 2/27, 0/1].
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-point-law · candidate
76. Conditional Candidate Eligibility and Signed Defect
S={0,…,1}, F=P(S), masses=[1, 1], D=2, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0], B=[0, 1], P(B)>0. Candidate=1/2. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
77. Conditional Candidate Eligibility and Signed Defect
S={0,…,1}, F=P(S), masses=[1, 1], D=2, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0], B=[0, 1], P(B)>0. Candidate=-1/1. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -3/2)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -3/2).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
78. Conditional Candidate Eligibility and Signed Defect
S={0,…,1}, F=P(S), masses=[1, 1], D=2, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0], B=[0, 1], P(B)>0. Candidate=2/1. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 3/2)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 3/2).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
79. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[3, 1, 2, 0], D=6, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=3/4. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
80. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[4, 1, 2, 0], D=7, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=5/5. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/5)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/5).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
81. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[5, 1, 2, 0], D=8, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=4/6. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/6)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/6).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
82. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[6, 1, 2, 0], D=9, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=6/7. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
83. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[7, 1, 2, 0], D=10, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=8/8. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/8)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/8).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
84. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[8, 1, 2, 0], D=11, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=7/9. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/9)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/9).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
85. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[9, 1, 2, 0], D=12, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=9/10. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
86. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[10, 1, 2, 0], D=13, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=11/11. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/11)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/11).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
87. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[11, 1, 2, 0], D=14, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=10/12. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/12)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/12).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
88. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[12, 1, 2, 0], D=15, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=12/13. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
89. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[13, 1, 2, 0], D=16, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=14/14. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/14)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/14).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
90. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[14, 1, 2, 0], D=17, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=13/15. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/15)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/15).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
91. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[15, 1, 2, 0], D=18, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=15/16. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
92. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[16, 1, 2, 0], D=19, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=17/17. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/17)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/17).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
93. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[17, 1, 2, 0], D=20, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=16/18. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/18)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/18).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
94. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[18, 1, 2, 0], D=21, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=18/19. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
95. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[19, 1, 2, 0], D=22, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=20/20. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/20)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/20).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
96. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[20, 1, 2, 0], D=23, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=19/21. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/21)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/21).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
97. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[21, 1, 2, 0], D=24, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=21/22. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
98. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[22, 1, 2, 0], D=25, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=23/23. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, 1/23)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, 1/23).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
99. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[23, 1, 2, 0], D=26, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=22/24. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(No, -1/24)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (No, -1/24).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
100. Conditional Candidate Eligibility and Signed Defect
S={0,…,3}, F=P(S), masses=[24, 1, 2, 0], D=27, and w_i=mass_i/D. Masses are nonnegative and sum exactly to D; uniform weights are not assumed. A=[0, 2], B=[0, 1], P(B)>0. Candidate=24/25. Give (does it equal P(A|B)?, candidate−P(A|B)). A candidate outside [0,1] may be submitted for rejection.
View answer and explanation
(Yes, 0/1)
Compute the actual conditional probability from the events, then subtract; only zero defect passes. An invalid probability candidate does not invalidate the underlying model. The exact result, retaining the requested order, is (Yes, 0/1).
Reviewed step-by-step derivation
Conditional probability in finite models · qf-density10-conditional-candidate-defect · candidate
Canonical question JSONL · English JSONL · Validation report