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A quadratic equation  
\[ax^2+bx+c=0\]
  will have real solutions if  
\[b^2-4ac \ge 0\]
. What is the probability that a randomly generated quadratic equation, with  
\[a, \: b, \: c\]
  all positive one digit numbers, will have solutions? All possible combinations of  
\[a, \: b, \: c\]
  returning real solutions are given in the table.
\[b\]
\[(a,c)\]
Number of possibilities
0
\[\begin{equation} \begin{aligned}(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0)\end{aligned} \end{equation}\]
9
1
\[\begin{equation} \begin{aligned}(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0)\end{aligned} \end{equation}\]
9
2
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \:(9,0), \\&(1,1)\end{aligned} \end{equation}\]
10
3
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1)\end{aligned} \end{equation}\]
12
4
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (3,1), \: (4,1), \: (2,2)\end{aligned} \end{equation}\]
17
5
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(3,1), \: (4,1), \\&(5,1), \: (6,1), \: (2,2), \: (2,3), \: (3,2)\end{aligned} \end{equation}\]
23
6
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \:(3,2), (4,2), \: (3,3)\end{aligned} \end{equation}\]
32
7
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \: (2,5), \: (2,6), \: (3,2), (4,2), \: (5,2), \: (6,2), \: (3,3), \\&(3,4), \: (4,3)\end{aligned} \end{equation}\]
38
8
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \: (2,5), \: (2,6), \: (2,7), \: (2,8), \: (3,2), \: (4,2), \: (5,2), \\&(6,2), \: (7,2), \: (8,2), \: (3,3), \: (3,4), \: (3,5), \: (4,3), \: (5,3), \: (4,4)\end{aligned} \end{equation}\]
44
9
\[\begin{equation} \begin{aligned}&(1,0), \: (2,0), \: (3,0), \: (4,0), \: (5,0), \: (6,0), \: (7,0), \: (8,0), \: (9,0), \\&(1,1), \: (1,2), \: (2,1), \: (1,3), \: (1,4), \: (1,5), \: (1,6), \:(1,7), \: (1,8), \\&(1,9), \: (3,1), \: (4,1), \: (5,1), \: (6,1), \: (7,1), \: (8,1), \: (9,1), \: (2,2), \\&(2,3), \: (2,4), \: (2,5), \: (2,6), \: (2,7), \: (2,8), \: (2,9), \: (3,2), \: (4,2), \\&(5,2), \: (6,2), \: (7,2), \: (8,2), \: (9,2), \:(3,3), \: (3,4), \: (3,5), \: (3,6), \\&(4,3), \: (5,3), \: (6,3), \: (4,4), \: (4,5), \: (5,4)\end{aligned} \end{equation}\]
51
There are 3245 combinations. There are 9 possible values for  
\[a\]
  (the digits 1 to 9) and 10 possibilities each for  
\[b\]
  and  
\[c\]
  so 900 possible values altogether. The probability is  
\[\frac{245}{900}=\frac{49}{180}\]
.