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प्रश्न
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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उत्तर
| 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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संबंधित प्रश्न
Which of the following can not be valid assignment of probabilities for outcomes of sample space S = {ω1, ω2,ω3,ω4,ω5,ω6,ω7}
| 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` |
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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 |
| (iv) |
\[\frac{1}{14}\]
|
\[\frac{2}{14}\]
|
\[\frac{3}{14}\]
|
\[\frac{4}{14}\]
|
\[\frac{5}{14}\]
|
\[\frac{6}{14}\]
|
\[\frac{15}{14}\]
|
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