The coefficients $a, b$ and $c$ of the quadratic equation, $a x^{2}+b x+c=0$ are obtained by throwing a dice three times. The probability that this equation has equal roots is

The quadratic equation $ax^{2} + bx + c = 0$ has equal roots if the discriminant $D = b^{2} - 4ac = 0$, which implies $b^{2} = 4ac$.

Since $a, b, c$ are outcomes of throwing a dice three times, $a, b, c \in \{1, 2, 3, 4, 5, 6\}$. The total number of possible outcomes is $6 \times 6 \times 6 = 216$.

We need to find the number of triplets $(a, b, c)$ such that $b^{2} = 4ac$.

If $b = 1$, $b^{2} = 1$, $4ac = 1$ (no integer solution).

If $b = 2$, $b^{2} = 4$, $4ac = 4 \Rightarrow ac = 1$. The only solution is $(a, c) = (1, 1)$. This gives 1 triplet: $(1, 2, 1)$.

If $b = 3$, $b^{2} = 9$, $4ac = 9$ (no integer solution).

If $b = 4$, $b^{2} = 16$, $4ac = 16 \Rightarrow ac = 4$. Possible pairs $(a, c)$ are $(1, 4), (2, 2), (4, 1)$. This gives 3 triplets: $(1, 4, 4), (2, 4, 2), (4, 4, 1)$.

If $b = 5$, $b^{2} = 25$, $4ac = 25$ (no integer solution).

If $b = 6$, $b^{2} = 36$, $4ac = 36 \Rightarrow ac = 9$. Possible pairs $(a, c)$ are $(3, 3)$. This gives 1 triplet: $(3, 6, 3)$.

The total number of favorable outcomes is $1 + 3 + 1 = 5$.

The probability is $\dfrac{5}{216}$.

Answer: $\dfrac{5}{216}$