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प्रश्न
If the letters of the word ASSASSINATION are arranged at random. Find the probability that no two A’s are coming together
उत्तर
Total number of word is ASSASSINATION are 13.
Where, we have 3A’s, 4S’, 2I’s, 2N’s, 1T’s and 1O’s.
If no two A’s are together, then arranging the alphabets except A’s
– S – S – S – S – I – N – T – I – O – N –
Number of ways of arranging all alphabets except A’s
= `(10!)/(4!2!2!)`
There are 11 vacant places between these alphabets.
∴ 3 A’s can be placed in 11 places in 11C3 ways
= `(11!)/(3!8!)`
∴ Total number of words when no two A’s together
= `(11!)/(3!8!) xx (101)/(4!2!2!)`
∴ Required probability = `(11! xx 10!)/(3!8!4!2!2!) xx (4!3!2!2!)/(13!)`
= `(10!)/(8! xx 13 xx 12)`
= `(10 xx 9 xx 8!)/(8! xx 13 xx 12)`
= `(10 xx 9)/(13 xx 12)`
= `15/26`
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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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\[\frac{1}{7}\]
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| (iv) |
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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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