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प्रश्न
If the letters of the word ASSASSINATION are arranged at random. Find the probability that four S’s come consecutively in the word
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 4 S’s come consecutively in the word, then arrangement may be as follows:
`(SSSS)/(1 "Group") (A A AI I N N T O)/(9 "others")`
∴ Number of words when all S’s are together = `(10!)/(3!2!21)` and the total number of word formed from the words
ASSASSINATION = `(13!)/(3!4!2!!)`
∴ Required probability = `((10!)/(3!2!2!))/((13!)/(3!4!2!2!))`
= `(10!)/(3!2!2!) xx (3!4!2!2!)/(13!)`
= `(10!4!)/(13!)`
= `(10! xx 4 xx 3 xx 2)/(13 xx 12 xx 11 xx 10!)`
= `2/143`
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संबंधित प्रश्न
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 |
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