Probability lies in [0,1]

$First\quad of\quad all,\quad the\quad total\quad number\quad of\quad outcomes\quad can\quad be\quad each\quad \\ number\quad with\quad all\quad others,\quad i.e.\quad (1,1);(1,2).....(10,9);(10,10)\quad \\ which\quad is\quad a\quad 100\quad outcomes.\\ \\ Now\quad for\quad { x }^{ 2 }+p.x+q=0\quad to\quad have\quad real\quad roots,\quad we\quad need\quad { p }^{ 2 }\ge 4q\\ So,\quad by\quad rejecting\quad cases\quad and\quad by\quad recognizing\quad a\quad pattern,\quad \\ We\quad can\quad see\quad for\quad p\quad =\quad 1,\quad q\quad has\quad no\quad value.\\ p=2,\quad q=1\\ p=3,\quad q=1,2\\ p=4,\quad q=1,2,3,4\\ p=5,\quad q=1,2,3,4,5,6\\ p=6,\quad q=1,2,3,4,5,6,7,8,9\\ p=7,8,9,10;\quad q=all\quad the\quad numbers.\\ So\quad total\quad number\quad of\quad pairs\quad possible=1+2+4+6+9+10+10+10+10\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =62\\ So,\quad probability\quad =\quad \frac { 62 }{ 100 } \quad =\quad 0.62$