1. Sample space

Each coefficient $a,b,c$ is obtained by throwing a fair die once, so each can be any of $1,2,3,4,5,6$.

Hence total number of possible ordered triples $(a,b,c)$ is $6^3=216.$

Since it is called a quadratic equation, we need $a\neq 0$, which is automatically true because a die never gives $0$.

2. Condition for equal roots

For the quadratic equation $ax^2+bx+c=0$ to have equal roots, its discriminant must be zero: $b^2-4ac=0.$ So we need $b^2=4ac.$

3. Check possible values of $b$

Since $b\in\{1,2,3,4,5,6\}$, we examine each value.

• If $b=1$, then $b^2=1$, so $4ac=1,$ impossible.

• If $b=2$, then $b^2=4$, so $4ac=4 \implies ac=1.$ This gives $(a,c)=(1,1).$

• If $b=3$, then $b^2=9$, so $4ac=9,$ impossible.

• If $b=4$, then $b^2=16$, so $4ac=16 \implies ac=4.$ Possible ordered pairs: $(a,c)=(1,4),(2,2),(4,1).$

• If $b=5$, then $b^2=25$, so $4ac=25,$ impossible.

• If $b=6$, then $b^2=36$, so $4ac=36 \implies ac=9.$ Possible ordered pairs: $(a,c)=(3,3).$

4. Count favorable cases

Thus favorable triples $(a,b,c)$ are:

• For $b=2$: $1$ case
• For $b=4$: $3$ cases
• For $b=6$: $1$ case

Total favorable cases: $1+3+1=5.$

5. Probability

Therefore, $P(\text{equal roots})=\frac{5}{216}.$

6. Compare with options

$\frac{5}{216}$ is Option B.