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प्रश्न
Two unbiased dice are thrown. Find the probability that neither a doublet nor a total of 8 will appear
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उत्तर
We know that in a single throw of two dices, the total number of possible outcomes is (6 × 6) = 36.
Let S be the sample space.
Then n(S) = 36
Let E1 = event of getting neither a doublet nor a total of 8
Then E1' = event of getting either a doublet or a total of 8
∴ E1' = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (2, 6), (3, 5), (5, 3), (6, 2)}
i.e. n(E1') = 10
Thus, P(E1') = \[\frac{10}{36}\]
Hence, required probability P(E1) = 1- P(E1')
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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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| (iv) |
\[\frac{1}{14}\]
|
\[\frac{2}{14}\]
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\[\frac{3}{14}\]
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\[\frac{4}{14}\]
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\[\frac{5}{14}\]
|
\[\frac{6}{14}\]
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\[\frac{15}{14}\]
|
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