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प्रश्न
A fair coin with 1 marked on one face and 6 on the other and a fair die are both tossed. Find the probability that the sum of numbers that turn up is (i) 3 (ii) 12
उत्तर
One die has 1 and 6 marked on it and the other has 1, 2, 3, 4, 5, 6
∴ Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(i) The sum of the given numbers is obtained by the event (1, 2).
Number of favourable outcomes = 1
∴ Probability when the sum of the numbers obtained is 3 = `1/12`
(ii) The sum of the given numbers is obtained by the event (6, 6). Here, number of favourable outcomes = 1
∴ Probability when the sum of the numbers obtained is 12 = `1/12`
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| 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 |
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| (iv) |
\[\frac{1}{14}\]
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\[\frac{2}{14}\]
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\[\frac{3}{14}\]
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\[\frac{4}{14}\]
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\[\frac{5}{14}\]
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\[\frac{6}{14}\]
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\[\frac{15}{14}\]
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