Вбанке 210 купюр одинакового достоинства распределены в равных количествах по трем банкоматам. среди этих купюр две юбилейные. какова вероятность того что обе юбилейные купюры окажутся в одном банкомате.

(12c+16d)-(6c-7d)=12с+16d-6c+7d=6c+23d
(11x^2-2x^2)-(x^3-x^2)=11x²-2x²-x³+x²=10x²-x³
(13x-11y+10z)-(-15x+10y-15z)=13x-11y+10z+15x-10y+15z=28x-21y+25z
(7m^2-4mn-n^2)-(2m^2-mn+2n^2)=7m²-4mn-n²-2m²+mn-2n²=5m²-3mn-3n²
(1/2x+2/3y-1/5z)-(-2/3x-1/2y+1/4z)=1/2x+2/3y-1/5z+2/3x+1/2y-1/4z=x+y-9/20z
(1/5ab+1/7bc-2/3ac)-(-4/5ab-3/14bc-1/5ac)=1/5ab+1/7bc-2/3ac+4/5ab+3/14bc+1/5ac=(1/5+4/5)ab+(1/7+3/14)bc+(1/5-2/3)ac=ab+5/14bc-7/15ac
(2/3x^3-3x^2y+1/4xy^2-2y^3-1)-(3x^3-2/3+1/2y^3-1/3x^2y-2xy^2)=2/3x³-3x²y+1/4xy²-2y³-1-3x³+2/3-1/2y³+1/3x²y+2xy²=(2/3-3)x³-x²y(3-1/3)+xy²(1/4+2)-y³(2+1/2)-1+2/3=-2 1/3x³-2 2/3x²y+2 1/4xy²-2 1/2y³-1/3
(0,6ab-0,5bc+cd)-(-0,5ab+2,5bc-cd)=0.6ab-0.5bc+cd+0.5ab-2.5bc+cd=1.1ab-3bc+2cd
(0,5abc+0,3bcd-1,5acd)-(-1,5abc+0,6bcd-2acd)=0.5abc+0.3bcd-1.5acd+1.5abc-0.6bcd+2acd=2abc-0.3bcd+0.5acd

(12c+16d)-(6c-7d)=12с+16d-6c+7d=6c+23d
(11x^2-2x^2)-(x^3-x^2)=11x²-2x²-x³+x²=10x²-x³
(13x-11y+10z)-(-15x+10y-15z)=13x-11y+10z+15x-10y+15z=28x-21y+25z
(7m^2-4mn-n^2)-(2m^2-mn+2n^2)=7m²-4mn-n²-2m²+mn-2n²=5m²-3mn-3n²
(1/2x+2/3y-1/5z)-(-2/3x-1/2y+1/4z)=1/2x+2/3y-1/5z+2/3x+1/2y-1/4z=x+y-9/20z
(1/5ab+1/7bc-2/3ac)-(-4/5ab-3/14bc-1/5ac)=1/5ab+1/7bc-2/3ac+4/5ab+3/14bc+1/5ac=(1/5+4/5)ab+(1/7+3/14)bc+(1/5-2/3)ac=ab+5/14bc-7/15ac
(2/3x^3-3x^2y+1/4xy^2-2y^3-1)-(3x^3-2/3+1/2y^3-1/3x^2y-2xy^2)=2/3x³-3x²y+1/4xy²-2y³-1-3x³+2/3-1/2y³+1/3x²y+2xy²=(2/3-3)x³-x²y(3-1/3)+xy²(1/4+2)-y³(2+1/2)-1+2/3=-2 1/3x³-2 2/3x²y+2 1/4xy²-2 1/2y³-1/3
(0,6ab-0,5bc+cd)-(-0,5ab+2,5bc-cd)=0.6ab-0.5bc+cd+0.5ab-2.5bc+cd=1.1ab-3bc+2cd
(0,5abc+0,3bcd-1,5acd)-(-1,5abc+0,6bcd-2acd)=0.5abc+0.3bcd-1.5acd+1.5abc-0.6bcd+2acd=2abc-0.3bcd+0.5acd