Finite Expectation Foundations
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batch-20261004-advance-20261005-r06-expectation-foundations · 2026-10-04 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
1. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -10), (1, -10), (2, -8), (3, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(6), 2/(6), 3/(6), 0/(6)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-9/1
The weighted numerator is -54 and the common denominator is 6, giving E[X]=-9/1. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
2. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-11/2
The weighted numerator is -44 and the common denominator is 8, giving E[X]=-11/2. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
3. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-6/1
The weighted numerator is -60 and the common denominator is 10, giving E[X]=-6/1. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
4. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-13/2
The weighted numerator is -78 and the common denominator is 12, giving E[X]=-13/2. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
5. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-21/5
The weighted numerator is -42 and the common denominator is 10, giving E[X]=-21/5. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
6. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-26/3
The weighted numerator is -104 and the common denominator is 12, giving E[X]=-26/3. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
7. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-8/7
The weighted numerator is -16 and the common denominator is 14, giving E[X]=-8/7. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
8. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-77/8
The weighted numerator is -154 and the common denominator is 16, giving E[X]=-77/8. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
9. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
1/7
The weighted numerator is 2 and the common denominator is 14, giving E[X]=1/7. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
10. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-49/4
The weighted numerator is -196 and the common denominator is 16, giving E[X]=-49/4. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
11. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
10/3
The weighted numerator is 60 and the common denominator is 18, giving E[X]=10/3. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
12. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-131/10
The weighted numerator is -262 and the common denominator is 20, giving E[X]=-131/10. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
13. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
13/3
The weighted numerator is 78 and the common denominator is 18, giving E[X]=13/3. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
14. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-16/1
The weighted numerator is -320 and the common denominator is 20, giving E[X]=-16/1. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
15. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
84/11
The weighted numerator is 168 and the common denominator is 22, giving E[X]=84/11. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
16. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-67/4
The weighted numerator is -402 and the common denominator is 24, giving E[X]=-67/4. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
17. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
93/11
The weighted numerator is 186 and the common denominator is 22, giving E[X]=93/11. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
18. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-119/6
The weighted numerator is -476 and the common denominator is 24, giving E[X]=-119/6. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
19. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
154/13
The weighted numerator is 308 and the common denominator is 26, giving E[X]=154/13. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
20. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-41/2
The weighted numerator is -574 and the common denominator is 28, giving E[X]=-41/2. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
21. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
163/13
The weighted numerator is 326 and the common denominator is 26, giving E[X]=163/13. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
22. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-166/7
The weighted numerator is -664 and the common denominator is 28, giving E[X]=-166/7. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
23. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
16/1
The weighted numerator is 480 and the common denominator is 30, giving E[X]=16/1. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
24. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
-389/16
The weighted numerator is -778 and the common denominator is 32, giving E[X]=-389/16. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
25. Expectation in a Nonuniform Finite Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.
View answer and explanation
83/5
The weighted numerator is 498 and the common denominator is 30, giving E[X]=83/5. Expectation may be negative or unattained and is not a sample-mean estimate.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-weighted · candidate
26. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -10), (1, -10), (2, -8), (3, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(6), 2/(6), 3/(6), 0/(6)], summing exactly to one. Let Z=(0)X+(-5). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
-5/1
Form Z at each outcome, then weight and sum to obtain -5/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
27. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Let Z=(-2)X+(-4). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
7/1
Form Z at each outcome, then weight and sum to obtain 7/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
28. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Let Z=(3)X+(-3). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
-21/1
Form Z at each outcome, then weight and sum to obtain -21/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
29. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Let Z=(-4)X+(-2). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
24/1
Form Z at each outcome, then weight and sum to obtain 24/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
30. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Let Z=(1)X+(-1). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
-26/5
Form Z at each outcome, then weight and sum to obtain -26/5, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
31. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Let Z=(-2)X+(0). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
52/3
Form Z at each outcome, then weight and sum to obtain 52/3, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
32. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Let Z=(3)X+(1). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
-17/7
Form Z at each outcome, then weight and sum to obtain -17/7, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
33. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Let Z=(-4)X+(2). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
81/2
Form Z at each outcome, then weight and sum to obtain 81/2, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
34. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Let Z=(1)X+(3). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
22/7
Form Z at each outcome, then weight and sum to obtain 22/7, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
35. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Let Z=(-2)X+(4). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
57/2
Form Z at each outcome, then weight and sum to obtain 57/2, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
36. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Let Z=(3)X+(5). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
15/1
Form Z at each outcome, then weight and sum to obtain 15/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
37. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Let Z=(-4)X+(6). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
292/5
Form Z at each outcome, then weight and sum to obtain 292/5, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
38. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Let Z=(1)X+(7). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
34/3
Form Z at each outcome, then weight and sum to obtain 34/3, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
39. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Let Z=(-2)X+(8). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
40/1
Form Z at each outcome, then weight and sum to obtain 40/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
40. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Let Z=(3)X+(9). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
351/11
Form Z at each outcome, then weight and sum to obtain 351/11, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
41. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Let Z=(-4)X+(10). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
77/1
Form Z at each outcome, then weight and sum to obtain 77/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
42. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Let Z=(1)X+(11). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
214/11
Form Z at each outcome, then weight and sum to obtain 214/11, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
43. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Let Z=(-2)X+(12). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
155/3
Form Z at each outcome, then weight and sum to obtain 155/3, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
44. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Let Z=(3)X+(13). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
631/13
Form Z at each outcome, then weight and sum to obtain 631/13, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
45. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Let Z=(-4)X+(14). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
96/1
Form Z at each outcome, then weight and sum to obtain 96/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
46. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Let Z=(1)X+(15). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
358/13
Form Z at each outcome, then weight and sum to obtain 358/13, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
47. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Let Z=(-2)X+(16). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
444/7
Form Z at each outcome, then weight and sum to obtain 444/7, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
48. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Let Z=(3)X+(17). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
65/1
Form Z at each outcome, then weight and sum to obtain 65/1, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
49. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Let Z=(-4)X+(18). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
461/4
Form Z at each outcome, then weight and sum to obtain 461/4, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
50. Expectation of an Affine Transformation
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Let Z=(1)X+(19). Find E[Z] exactly as a reduced fraction with positive denominator.
View answer and explanation
178/5
Form Z at each outcome, then weight and sum to obtain 178/5, equivalently aE[X]+b. At a=0 only the constant b remains.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-affine · candidate
51. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -10), (1, -10), (2, -8), (3, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(6), 2/(6), 3/(6), 0/(6)], summing exactly to one. On the same S, Y has table {(0, -20), (1, -20), (2, -16), (3, 1998)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-27/1
At each identical outcome, add X and Y first and then use the declared weights, giving -27/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
52. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. On the same S, Y has table {(0, -13), (1, -13), (2, -7), (3, 2001)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-31/2
At each identical outcome, add X and Y first and then use the declared weights, giving -31/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
53. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. On the same S, Y has table {(0, -14), (1, -14), (2, -6), (3, 2004)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-16/1
At each identical outcome, add X and Y first and then use the declared weights, giving -16/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
54. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. On the same S, Y has table {(0, -15), (1, -15), (2, -5), (3, 2007)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-33/2
At each identical outcome, add X and Y first and then use the declared weights, giving -33/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
55. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. On the same S, Y has table {(0, -8), (1, -8), (2, 4), (3, 2010)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-43/5
At each identical outcome, add X and Y first and then use the declared weights, giving -43/5. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
56. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. On the same S, Y has table {(0, -22), (1, -22), (2, -8), (3, 2008)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-26/1
At each identical outcome, add X and Y first and then use the declared weights, giving -26/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
57. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. On the same S, Y has table {(0, -7), (1, -7), (2, 9), (3, 2011)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-17/7
At each identical outcome, add X and Y first and then use the declared weights, giving -17/7. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
58. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. On the same S, Y has table {(0, -24), (1, -24), (2, -6), (3, 2014)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-215/8
At each identical outcome, add X and Y first and then use the declared weights, giving -215/8. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
59. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. On the same S, Y has table {(0, -1), (1, -1), (2, 19), (3, 2017)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
24/7
At each identical outcome, add X and Y first and then use the declared weights, giving 24/7. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
60. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. On the same S, Y has table {(0, -26), (1, -26), (2, -4), (3, 2020)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-131/4
At each identical outcome, add X and Y first and then use the declared weights, giving -131/4. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
61. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. On the same S, Y has table {(0, 0), (1, 0), (2, 24), (3, 2018)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
10/1
At each identical outcome, add X and Y first and then use the declared weights, giving 10/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
62. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. On the same S, Y has table {(0, -33), (1, -33), (2, -7), (3, 2021)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-383/10
At each identical outcome, add X and Y first and then use the declared weights, giving -383/10. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
63. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. On the same S, Y has table {(0, 6), (1, 6), (2, 34), (3, 2024)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
15/1
At each identical outcome, add X and Y first and then use the declared weights, giving 15/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
64. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. On the same S, Y has table {(0, -35), (1, -35), (2, -5), (3, 2027)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-45/1
At each identical outcome, add X and Y first and then use the declared weights, giving -45/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
65. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. On the same S, Y has table {(0, 12), (1, 12), (2, 44), (3, 2030)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
296/11
At each identical outcome, add X and Y first and then use the declared weights, giving 296/11. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
66. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. On the same S, Y has table {(0, -42), (1, -42), (2, -8), (3, 2028)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-201/4
At each identical outcome, add X and Y first and then use the declared weights, giving -201/4. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
67. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. On the same S, Y has table {(0, 13), (1, 13), (2, 49), (3, 2031)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
290/11
At each identical outcome, add X and Y first and then use the declared weights, giving 290/11. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
68. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. On the same S, Y has table {(0, -44), (1, -44), (2, -6), (3, 2034)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-115/2
At each identical outcome, add X and Y first and then use the declared weights, giving -115/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
69. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. On the same S, Y has table {(0, 19), (1, 19), (2, 59), (3, 2037)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
501/13
At each identical outcome, add X and Y first and then use the declared weights, giving 501/13. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
70. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. On the same S, Y has table {(0, -46), (1, -46), (2, -4), (3, 2040)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-115/2
At each identical outcome, add X and Y first and then use the declared weights, giving -115/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
71. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. On the same S, Y has table {(0, 20), (1, 20), (2, 64), (3, 2038)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
489/13
At each identical outcome, add X and Y first and then use the declared weights, giving 489/13. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
72. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. On the same S, Y has table {(0, -53), (1, -53), (2, -7), (3, 2041)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-491/7
At each identical outcome, add X and Y first and then use the declared weights, giving -491/7. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
73. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. On the same S, Y has table {(0, 26), (1, 26), (2, 74), (3, 2044)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
50/1
At each identical outcome, add X and Y first and then use the declared weights, giving 50/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
74. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. On the same S, Y has table {(0, -55), (1, -55), (2, -5), (3, 2047)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
-1119/16
At each identical outcome, add X and Y first and then use the declared weights, giving -1119/16. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
75. Expectation of a Sum on the Same Model
The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. On the same S, Y has table {(0, 32), (1, 32), (2, 84), (3, 2050)}. Compute E[X+Y] exactly without introducing a separate independent model.
View answer and explanation
269/5
At each identical outcome, add X and Y first and then use the declared weights, giving 269/5. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-same-model-sum · candidate
76. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-12 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-12/1
The constant multiplies the weight total one, giving E[X]=-12/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
77. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-11 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-11/1
The constant multiplies the weight total one, giving E[X]=-11/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
78. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-10 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-10/1
The constant multiplies the weight total one, giving E[X]=-10/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
79. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-9 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-9/1
The constant multiplies the weight total one, giving E[X]=-9/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
80. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-8 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-8/1
The constant multiplies the weight total one, giving E[X]=-8/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
81. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-7 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-7/1
The constant multiplies the weight total one, giving E[X]=-7/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
82. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-6 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-6/1
The constant multiplies the weight total one, giving E[X]=-6/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
83. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-5 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-5/1
The constant multiplies the weight total one, giving E[X]=-5/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
84. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-4 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-4/1
The constant multiplies the weight total one, giving E[X]=-4/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
85. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-3 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-3/1
The constant multiplies the weight total one, giving E[X]=-3/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
86. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-2 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-2/1
The constant multiplies the weight total one, giving E[X]=-2/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
87. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-1 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
-1/1
The constant multiplies the weight total one, giving E[X]=-1/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
88. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=0 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
0/1
The constant multiplies the weight total one, giving E[X]=0/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
89. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=1 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
1/1
The constant multiplies the weight total one, giving E[X]=1/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
90. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=2 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
2/1
The constant multiplies the weight total one, giving E[X]=2/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
91. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=3 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
3/1
The constant multiplies the weight total one, giving E[X]=3/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
92. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=4 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
4/1
The constant multiplies the weight total one, giving E[X]=4/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
93. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=5 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
5/1
The constant multiplies the weight total one, giving E[X]=5/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
94. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=6 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
6/1
The constant multiplies the weight total one, giving E[X]=6/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
95. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=7 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
7/1
The constant multiplies the weight total one, giving E[X]=7/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
96. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=8 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
8/1
The constant multiplies the weight total one, giving E[X]=8/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
97. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=9 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
9/1
The constant multiplies the weight total one, giving E[X]=9/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
98. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=10 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
10/1
The constant multiplies the weight total one, giving E[X]=10/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
99. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=11 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
11/1
The constant multiplies the weight total one, giving E[X]=11/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
100. Expectation of a Constant in a Normalized Model
The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=12 at every outcome, compute E[X] as a reduced fraction with positive denominator.
View answer and explanation
12/1
The constant multiplies the weight total one, giving E[X]=12/1. Different normalized weight representations do not change this constant-expectation result.
Reviewed step-by-step derivation
Expectation in finite models · qf-expectation-constant · candidate
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