Probability of randomly generated quadratic equation having equal roots Could any one help me to solve this problem?
Given that the coefficients of the equation $ax^2+bx+c=0$ are selected by throwing an unbiased die, we need to find what is the probability of the equation having equal roots.
Thank you for  Hints.
 A: Hint: The roots are the same iff $b^2-4ac=0$ so find the number of triples $(a,b,c)$ with each of $a,b,c$ in $\{1,2,3,4,5,6\}$ for which this equation holds, then divide by $6^3$.
OP has asked about finding the $a,b,c$: First from $b^2=4ac$ we see that $b$ must be even, so that it is one of $2,4,6$
$b=2$ (so that $b^2=4=4\cdot 1$) implies $ac=1$ giving the single pair $(1,1)$ for $(a,c).$
$b=4$ (so that $b^2=16=4\cdot 4$) implies $ac=4$ giving the three pairs $(1,4),(2,2),(4,1)$ for $(a,c)$
Finally $b=6$ (so that $b^2=36=4\cdot 9$ implies $ac=9$ giving one pair $(3,3)$ for $(a,b)$. [we can't have $(1,9),(9,1)$ since dice only go to $6$]
So the list of triples $(a,b,c)$ for which $b^2=4ac$ is 
$$(1,2,1),(1,4,4),(2,4,2),(4,4,1),(3,6,3).$$
A: for this first you have to find the maximum number of possibility i.e
6*6*6 = 6^3
for having equal roots:
b^2 = 4ac
so it means that the value of 4ac must be equal to {1, 4, 9, 16, 25, 36}
so the possibilities of a and c to take the values are 1 for (1,1) & 1 for (2,2) &1 for (3,3) and for different 1 for (1,4), 1 for (4,1) so in sum gives = 5 possibilities
so 5/6^3 according to me is the answer...
