Question -
| Assignment | ω1 | ω2 | ω3 | ω4 | ω5 | ω6 | ω7 |
| (a) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
| (b) | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 | 1/7 |
| (c) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
| (d) | -0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
| (e) | 1/14 | 2/14 | 3/14 | 4/14 | 5/14 | 6/14 | 15/14 |
Answer -
(a) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Therefore, the given assignment is valid.
b) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
= (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7) + (1/7)
= 7/7
= 1
Therefore, the given assignment is valid.
c) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7
= 2.8 > 1
Therefore, the 2nd condition is notsatisfied
Which states that p(wi) ≤ 1
So, the given assignment is not valid.
d) The conditions of axiomatic approach don’t hold true inthe given assignment, that is
1) Each of the number p(wi) is less than zero butalso negative
To be true each of the number p(wi) should beless than zero and positive
So, the assignment is not valid
e) Condition (i): Each of the number p(ωi )is positive and less than zero. Condition (ii): Sum of probabilities
= (1/14) + (2/14) + (3/14) + (4/14) + (5/14) + (6/14) +(7/14)
= (28/14) ≥ 1
The second condition doesn’t hold true so the assignment isnot valid.