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प्रश्न
6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is ______.
पर्याय
`1/432`
`12/431`
`1/132`
None of these
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उत्तर
6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is `1/132`.
Explanation:
If all the girls sit together, then we consider it as 1 group
∴ Total number of arrangement of 6 + 1 = 7 persons in a row
= 7! and the girls also interchanged their places with 6! ways.
∴ Required probability = `(6!7!)/(12!)`
= `(6 xx 5 xx 4 xx 3 xx 2 xx 7!)/(12 xx 11 xx 10 xx 9 xx 8 xx 7!)`
= `1/132`.
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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 |
| (e) | `1/14` | `2/14` | `3/14` | `4/14` | `5/14` | `6/14` | `15/14` |
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| (i) | 0.1 | 0.01 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
Which of the cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w1, w2, w3, w4, w5, w6, w7}:
| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (ii) |
\[\frac{1}{7}\]
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\[\frac{1}{7}\]
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\[\frac{1}{7}\]
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\[\frac{1}{7}\]
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\[\frac{1}{7}\]
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\[\frac{1}{7}\]
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\[\frac{1}{7}\]
|
Which of the cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w1, w2, w3, w4, w5, w6, w7}:
| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (iii) | 0.7 | 0.06 | 0.05 | 0.04 | 0.03 | 0.2 | 0.1 |
Which of the cannot be valid assignment of probability for elementary events or outcomes of sample space S = {w1, w2, w3, w4, w5, w6, w7}:
| Elementary events: | w1 | w2 | w3 | w4 | w5 | w6 | w7 |
| (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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