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Question
Two dice are thrown together. The probability that neither they show equal digits nor the sum of their digits is 9 will be
Options
13/15
13/18
1/9
8/9
Solution
13/18
When two dices are thrown, there are (6 × 6) = 36 outcomes.
The set of all these outcomes is the sample space is given by
S = (1, 1) , (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)
(2, 1) , (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)
(3, 1) , (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)
(4, 1) , (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)
(5, 1) , (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)
(6, 1) , (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)
∴ n(S) = 36
Let E be the event of getting the digits which are neither equal nor give a total of 9.
Then E' = event of getting either a doublet or a total of 9
Thus, E' = {{1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (3, 6), (4, 5), (5, 4), (6, 3)}
i.e. n(E') = 10
P(E') = \[\frac{n\left( E' \right)}{n\left( S \right)} = \frac{10}{36} = \frac{5}{18}\]
Hence, required probability P(E) = 1- P(E')
= \[1 - \frac{5}{18} = \frac{13}{18}\]
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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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