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प्रश्न
If the letters of the word ASSASSINATION are arranged at random. Find the probability that all A’s are not coming together
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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 all A’s are coming together, then three are 11 alphabets
Number of words when all A’s come together
= `(11!)/(4!2!2!)`
∴ Probability when all A’s come together
= `((11!)/(4!2!2!))/((13!)/(4!3!2!2!))`
= `(11!)/(4!2!2!) xx (4!3!2!2!)/(13!)`
= `(11! xx 3!)/(13!)`
= `6/(13 xx 12)`
= `1/26`
∴ Required probability when all A’s do not come together
= `1 - 1/26 = 25/26`
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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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