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network profile
| bio | website | |
|---|---|---|
| location | ||
| age | ||
| visits | member for | 6 months |
| seen | May 10 at 23:04 | |
| stats | profile views | 5 |
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May 11 |
awarded | Announcer |
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Jan 5 |
awarded | Supporter |
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Jan 5 |
accepted | Conditional probability exercise |
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Jan 5 |
comment |
Conditional probability exercise Thank you, I assumed I am looking for Pr(P|N) which was the wrong way. |
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Jan 5 |
asked | Conditional probability exercise |
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Oct 26 |
awarded | Editor |
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Oct 26 |
revised |
Number of surjective functions$ f: A->B$ where $f(1) > f(2) > f(3)$ edited body |
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Oct 26 |
awarded | Student |
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Oct 26 |
awarded | Custodian |
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Oct 26 |
reviewed | Approve suggested edit on Number of surjective functions$ f: A->B$ where $f(1) > f(2) > f(3)$ |
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Oct 26 |
awarded | Scholar |
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Oct 26 |
accepted | Number of surjective functions$ f: A->B$ where $f(1) > f(2) > f(3)$ |
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Oct 26 |
comment |
Number of surjective functions$ f: A->B$ where $f(1) > f(2) > f(3)$ I see! It is 4 * (4^3 - 3^3). 4^3 is number of ways how to map remaining els MINUS the ways how to map remaining elements without mapping to the one element of B. |
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Oct 26 |
asked | Number of surjective functions$ f: A->B$ where $f(1) > f(2) > f(3)$ |
