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Question
Which of the cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w1, w2, w3, w4, w5, w6, w7}:
| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (ii) |
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
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Solution
| w1 | w2 | w3 | w4 | w5 | w6 | w7 |
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
\[\frac{1}{7}\]
|
Here, each of the numbers p(ωi) is positive and less than 1.
∴ Sum of probabilities = \[p\left( \omega_1 \right) + p\left( \omega_2 \right) + p\left( \omega_3 \right) + p\left( \omega_4 \right) + p\left( \omega_5 \right) + p\left( \omega_6 \right) + p\left( \omega_7 \right)\]
= \[\frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} + \frac{1}{7} = 7 \times \frac{1}{7} = 1\]
Thus, the assignment is valid.
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