Finite Probability Foundations
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This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261003-r07-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. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-12, -11} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
0/1
Explicit equiprobability permits |E|/|S|=0/2=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
2. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-11, -10, -9}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/1
Explicit equiprobability permits |E|/|S|=3/3=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
3. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-10, -9, -8, -7} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-10}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/4
Explicit equiprobability permits |E|/|S|=1/4=1/4. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
4. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-9, -8, -7, -6, -5} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-5}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/5
Explicit equiprobability permits |E|/|S|=1/5=1/5. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
5. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-8, -7, -6, -5, -4, -3} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-8, -6, -4}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/2
Explicit equiprobability permits |E|/|S|=3/6=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
6. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-7, -6} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-7, -6}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/1
Explicit equiprobability permits |E|/|S|=2/2=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
7. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
0/1
Explicit equiprobability permits |E|/|S|=0/3=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
8. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-5, -4, -3, -2} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-5, -4, -3, -2}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/1
Explicit equiprobability permits |E|/|S|=4/4=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
9. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-4, -3, -2, -1, 0} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-4}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/5
Explicit equiprobability permits |E|/|S|=1/5=1/5. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
10. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-3, -2, -1, 0, 1, 2} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={2}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/6
Explicit equiprobability permits |E|/|S|=1/6=1/6. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
11. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-2, -1} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-2}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/2
Explicit equiprobability permits |E|/|S|=1/2=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
12. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-1, 0}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
2/3
Explicit equiprobability permits |E|/|S|=2/3=2/3. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
13. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={0, 1, 2, 3} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
0/1
Explicit equiprobability permits |E|/|S|=0/4=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
14. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={1, 2, 3, 4, 5} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={1, 2, 3, 4, 5}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/1
Explicit equiprobability permits |E|/|S|=5/5=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
15. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={2, 3, 4, 5, 6, 7} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={2}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/6
Explicit equiprobability permits |E|/|S|=1/6=1/6. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
16. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={3, 4} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={4}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/2
Explicit equiprobability permits |E|/|S|=1/2=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
17. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={4, 6}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
2/3
Explicit equiprobability permits |E|/|S|=2/3=2/3. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
18. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={5, 6, 7, 8} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={5, 6}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/2
Explicit equiprobability permits |E|/|S|=2/4=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
19. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={6, 7, 8, 9, 10} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
0/1
Explicit equiprobability permits |E|/|S|=0/5=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
20. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={7, 8, 9, 10, 11, 12} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={7, 8, 9, 10, 11, 12}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/1
Explicit equiprobability permits |E|/|S|=6/6=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
21. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={8, 9} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={8}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/2
Explicit equiprobability permits |E|/|S|=1/2=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
22. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={11}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/3
Explicit equiprobability permits |E|/|S|=1/3=1/3. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
23. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={10, 11, 12, 13} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={10, 12}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
1/2
Explicit equiprobability permits |E|/|S|=2/4=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
24. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={11, 12, 13, 14, 15} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={11, 12}, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
2/5
Explicit equiprobability permits |E|/|S|=2/5=2/5. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
25. Event Probability with Explicit Equiprobability
The nonempty finite sample space is S={12, 13, 14, 15, 16, 17} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.
View answer and explanation
0/1
Explicit equiprobability permits |E|/|S|=0/6=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-uniform-event · candidate
26. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-12, -11, -10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(2), 1/(2), 0/(2)] and E={-10}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
27. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(5), 2/(5), 1/(5)] and E={-11, -10, -9}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
28. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-10, -9, -8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(8), 3/(8), 2/(8)] and E={-10}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
3/8
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 3/8. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
29. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-9, -8, -7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [4/(8), 4/(8), 0/(8)] and E={-7}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
30. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-8, -7, -6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [5/(11), 5/(11), 1/(11)] and E={-8, -7}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
10/11
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 10/11. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
31. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-7, -6, -5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [6/(9), 1/(9), 2/(9)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
32. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [7/(9), 2/(9), 0/(9)] and E={-6, -5, -4}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
33. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-5, -4, -3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [8/(12), 3/(12), 1/(12)] and E={-5}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
2/3
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 2/3. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
34. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-4, -3, -2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [9/(15), 4/(15), 2/(15)] and E={-2}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
2/15
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 2/15. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
35. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-3, -2, -1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [10/(15), 5/(15), 0/(15)] and E={-3, -2}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
36. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-2, -1, 0} and the event family is F=𝒫(S). In the displayed order of S, the weights are [11/(13), 1/(13), 1/(13)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
37. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [12/(16), 2/(16), 2/(16)] and E={-1, 0, 1}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
38. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={0, 1, 2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [13/(16), 3/(16), 0/(16)] and E={0}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
13/16
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 13/16. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
39. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={1, 2, 3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [14/(19), 4/(19), 1/(19)] and E={3}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/19
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/19. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
40. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={2, 3, 4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [15/(22), 5/(22), 2/(22)] and E={2, 3}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
10/11
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 10/11. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
41. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={3, 4, 5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [16/(17), 1/(17), 0/(17)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
42. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [17/(20), 2/(20), 1/(20)] and E={4, 5, 6}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
43. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={5, 6, 7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [18/(23), 3/(23), 2/(23)] and E={5}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
18/23
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 18/23. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
44. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={6, 7, 8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [19/(23), 4/(23), 0/(23)] and E={8}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
45. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={7, 8, 9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [20/(26), 5/(26), 1/(26)] and E={7, 8}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
25/26
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 25/26. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
46. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={8, 9, 10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [21/(24), 1/(24), 2/(24)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
47. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). In the displayed order of S, the weights are [22/(24), 2/(24), 0/(24)] and E={9, 10, 11}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
48. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={10, 11, 12} and the event family is F=𝒫(S). In the displayed order of S, the weights are [23/(27), 3/(27), 1/(27)] and E={10}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
23/27
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 23/27. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
49. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={11, 12, 13} and the event family is F=𝒫(S). In the displayed order of S, the weights are [24/(30), 4/(30), 2/(30)] and E={13}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/15
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/15. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
50. Weighted Event Probability without Equiprobability
The nonempty finite sample space is S={12, 13, 14} and the event family is F=𝒫(S). In the displayed order of S, the weights are [25/(30), 5/(30), 0/(30)] and E={12, 13}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-weighted-event · candidate
51. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-12, -11, -10} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [1/(2), 1/(2), 0/(2)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 2 and the common denominator is 2. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
52. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-2/(5), 2/(5), 1/(5)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 1 and the common denominator is 5. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
53. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-10, -9, -8} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [3/(9), 3/(9), 2/(9)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 8 and the common denominator is 9. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
54. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-9, -8, -7} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [4/(0), 4/(0), 0/(0)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 8 and the common denominator is 0. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
55. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-8, -7, -6} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [5/(11), 5/(11), 1/(11)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 11 and the common denominator is 11. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
56. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-7, -6, -5} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-6/(9), 1/(9), 2/(9)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to -3 and the common denominator is 9. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
57. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [7/(10), 2/(10), 0/(10)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 9 and the common denominator is 10. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
58. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-5, -4, -3} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [8/(-12), 3/(-12), 1/(-12)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 12 and the common denominator is -12. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
59. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-4, -3, -2} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [9/(15), 4/(15), 2/(15)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 15 and the common denominator is 15. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
60. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-3, -2, -1} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-10/(15), 5/(15), 0/(15)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to -5 and the common denominator is 15. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
61. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-2, -1, 0} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [11/(14), 1/(14), 1/(14)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 13 and the common denominator is 14. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
62. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [12/(-16), 2/(-16), 2/(-16)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 16 and the common denominator is -16. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
63. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={0, 1, 2} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [13/(16), 3/(16), 0/(16)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 16 and the common denominator is 16. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
64. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={1, 2, 3} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-14/(19), 4/(19), 1/(19)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to -9 and the common denominator is 19. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
65. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={2, 3, 4} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [15/(23), 5/(23), 2/(23)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 22 and the common denominator is 23. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
66. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={3, 4, 5} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [16/(-17), 1/(-17), 0/(-17)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 17 and the common denominator is -17. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
67. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [17/(20), 2/(20), 1/(20)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 20 and the common denominator is 20. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
68. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={5, 6, 7} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-18/(23), 3/(23), 2/(23)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to -13 and the common denominator is 23. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
69. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={6, 7, 8} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [19/(24), 4/(24), 0/(24)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 23 and the common denominator is 24. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
70. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={7, 8, 9} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [20/(-26), 5/(-26), 1/(-26)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 26 and the common denominator is -26. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
71. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={8, 9, 10} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [21/(24), 1/(24), 2/(24)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 24 and the common denominator is 24. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
72. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-22/(24), 2/(24), 0/(24)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to -20 and the common denominator is 24. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
73. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={10, 11, 12} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [23/(28), 3/(28), 1/(28)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 27 and the common denominator is 28. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
74. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={11, 12, 13} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [24/(-30), 4/(-30), 2/(-30)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
No
The weight numerators sum to 30 and the common denominator is -30. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
75. Validity of a Candidate Finite Probability Model
The nonempty finite sample space is S={12, 13, 14} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [25/(30), 5/(30), 0/(30)]. Do they define a valid probability model with nonnegative weights summing to one?
View answer and explanation
Yes
The weight numerators sum to 30 and the common denominator is 30. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-model-validity · candidate
76. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-12, -11, -10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(3), 1/(3), 2/(3)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
77. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(6), 2/(6), 3/(6)] and E={-11, -10, -9}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
78. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-10, -9, -8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(9), 3/(9), 4/(9)] and E={-10}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
7/9
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 7/9. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
79. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-9, -8, -7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(12), 4/(12), 5/(12)] and E={-7}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
7/12
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 7/12. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
80. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-8, -7, -6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(11), 5/(11), 6/(11)] and E={-8, -7}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
6/11
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 6/11. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
81. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-7, -6, -5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(9), 1/(9), 7/(9)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
82. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(12), 2/(12), 8/(12)] and E={-6, -5, -4}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
83. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-5, -4, -3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(15), 3/(15), 9/(15)] and E={-5}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
4/5
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 4/5. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
84. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-4, -3, -2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(14), 4/(14), 10/(14)] and E={-2}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
2/7
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 2/7. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
85. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-3, -2, -1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(17), 5/(17), 11/(17)] and E={-3, -2}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
11/17
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 11/17. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
86. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-2, -1, 0} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(15), 1/(15), 12/(15)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
87. Complement Probability in a Weighted Model
The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(18), 2/(18), 13/(18)] and E={-1, 0, 1}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
0/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
88. Complement Probability in a Weighted Model
The nonempty finite sample space is S={0, 1, 2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(17), 3/(17), 14/(17)] and E={0}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
89. Complement Probability in a Weighted Model
The nonempty finite sample space is S={1, 2, 3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(20), 4/(20), 15/(20)] and E={3}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
View answer and explanation
1/4
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/4. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
90. Complement Probability in a Weighted Model
The nonempty finite sample space is S={2, 3, 4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(23), 5/(23), 16/(23)] and E={2, 3}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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16/23
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 16/23. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
91. Complement Probability in a Weighted Model
The nonempty finite sample space is S={3, 4, 5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(21), 1/(21), 17/(21)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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1/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
92. Complement Probability in a Weighted Model
The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(20), 2/(20), 18/(20)] and E={4, 5, 6}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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0/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
93. Complement Probability in a Weighted Model
The nonempty finite sample space is S={5, 6, 7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(23), 3/(23), 19/(23)] and E={5}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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22/23
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 22/23. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
94. Complement Probability in a Weighted Model
The nonempty finite sample space is S={6, 7, 8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(26), 4/(26), 20/(26)] and E={8}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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3/13
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 3/13. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
95. Complement Probability in a Weighted Model
The nonempty finite sample space is S={7, 8, 9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(29), 5/(29), 21/(29)] and E={7, 8}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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21/29
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 21/29. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
96. Complement Probability in a Weighted Model
The nonempty finite sample space is S={8, 9, 10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(23), 1/(23), 22/(23)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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1/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
97. Complement Probability in a Weighted Model
The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(26), 2/(26), 23/(26)] and E={9, 10, 11}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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0/1
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
98. Complement Probability in a Weighted Model
The nonempty finite sample space is S={10, 11, 12} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(29), 3/(29), 24/(29)] and E={10}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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27/29
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 27/29. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
99. Complement Probability in a Weighted Model
The nonempty finite sample space is S={11, 12, 13} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(32), 4/(32), 25/(32)] and E={13}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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7/32
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 7/32. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
100. Complement Probability in a Weighted Model
The nonempty finite sample space is S={12, 13, 14} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(31), 5/(31), 26/(31)] and E={12, 13}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.
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26/31
Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 26/31. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law.
Reviewed step-by-step derivation
Probability on finite sample spaces · qf-probability-complement · candidate
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