A pair of fair dice is thrown independently three times. The probability of getting a score of exactly 9 twice is

The outcome for a pair of fair dice can be represented as the ordered pair $$(x,y)$$ with $$x,y \in \{1,2,\dots,6\}$$, giving $$6 \times 6 = 36$$ equally likely results.

Step 1: Favourable cases for a score of 9.
A sum of $$9$$ occurs for the pairs $$(3,6), (4,5), (5,4), (6,3)$$—exactly $$4$$ outcomes.

Hence, for a single throw of the two dice,

$$P(\text{score } 9) = \frac{4}{36} = \frac{1}{9}, \qquad P(\text{score }\neq 9) = 1 - \frac{1}{9} = \frac{8}{9}.$$

Step 2: Model the three independent throws.
Let “success’’ mean “score is 9’’. We need exactly two successes in three trials, which follows the binomial distribution:

$$P(\text{exactly } 2 \text{ scores of } 9) = \binom{3}{2}\left(\frac{1}{9}\right)^{2}\left(\frac{8}{9}\right).$$

Calculate:

$$\binom{3}{2} = 3,\quad \left(\frac{1}{9}\right)^{2} = \frac{1}{81},\quad 3 \times \frac{1}{81} \times \frac{8}{9} = 3 \times \frac{8}{729} = \frac{24}{729}.$$

Simplify $$\frac{24}{729}$$ by dividing numerator and denominator by $$3$$:

$$\frac{24}{729} = \frac{8}{243}.$$

Therefore, the required probability is $$\frac{8}{243}$$.

Option D which is: $$\dfrac{8}{243}$$