3-Mustaqil ish Noaniqliklarni taqdim etish uchun shartli ehtimollik: afzallik va kamchiliklari Reja


Noravshan to`plamlarning dekart ko`paytmasi


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Mustaqil ish 3

Noravshan to`plamlarning dekart ko`paytmasi. Ai,i1,n NTostilarning dekart ko`paytmasi-bu
A1A2An{(x(x1,x2,xn)/(x1,x2,xn))}
to`plami, bu yerda xi  Xi; x (x1, x2,, xn )  min{ A1 (x1), A2 (x2 ),,An (xn )}.
Misol.Х={10,15,20,25}vaY={5,6,7}ko`rinishdagi asosiy
to`plamlar berilgan. Ushbu to`plamlarda noravshan top’lamostilari
A1{,,,} va A2 {, < 0.5/6 >, < 0.2/7 >}
belgilangan. Ular ustida dekart ko`paytmasi amali bajarish natijasida
quyidagi natijalar olinadi:
< 0.2/(10.7) >, < 0.2/(15,7) >, < 0.2/(20,7) >, < 0.2/(25,7) >};
< 0.5/(10.6) >, < 0.5/(15,6) >, < 0.5/(20,6) >, < 0.3/(25,6) >,
A1  A2 = {< 1/(10,5) >, < 0.8/(15,5) >, < 0.5/(20,5) >, < 0.3/(25.5) >,
NoTlarning ayirmasi. U to‘plamdagi ikkita A va B NTostilarining
ayirmasi - bu A \ B  { A\B (x)  / x}, x  X, bu yerda A\B (x)  A(x)  B (x)
ko`rinishdagi to`plam.
Misol. Aytaylik,
B (x)  1 B (x) asosan B  B(x) 1 B(x)  (x1 | 0.65),(x2 | 0.13),(x3 |1),(x4 | 0)
Endi A\ B { A\B (x)  / x} A\B (x)  A(x)  B (x)  min(A(x)  B (x))
asosan A\B(x1|0.25),(x2|0.13),(x3|1),(x4|0).
NoTlarning simmetrik ayirmasi. U to‘plamdagi ikkita A va B
NoTostilarining simmetrik ayirmasi - bu AB { AB (x)  / x}, x X ,
bu yerda AB (x)  A\B (x)  B\ A (x) ko`rinishdagi to`plam.
Misol. Aytaylik,
1) B (x) 1 B (x) asosan B  B (x) 1 B (x)  (x1 | 0.65),(x2 | 0.13),(x3 |1),(x4 | 0). Endi A\ B { A\B (x)  / x}  A\B (x)  A(x)  B (x)  min(A(x), B (x))
asosan A\B (x)  A \ B  (x1 | 0.25),(x2 | 0.13),(x3 |1),(x4 | 0).
2) A (x) 1 A (x) asosan A  A (x) 1 A (x)  (x1 | 0.75),(x2 | 0.27),(x3 | 0),(x4 |1). Endi B \ A { B\ A(x)  / x}  B\ A(x)  B (x)  A(x)  min(B (x), A(x))
asosan B\ A(x)  B \ A  (x1 | 0.35),(x2 | 0.27),(x3 | 0),(x4 |1).
3) AB (x)  A\B (x)  B\ A(x) va A(x)  B (x)  max(A(x),B (x))
formulalarga asosan AB (x)  A\B (x)  B\ A (x)  (x1 | 0.35),(x2 | 0.27),(x3 |1),(x4 |1).

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