Two dices are rolled. If both dices have six faces numbered 1, 2, 3, 5, 7 and 11, then the probability that the sum of the numbers on the top faces is less than or equal to 8 is:

Both dice have faces numbered $$1, 2, 3, 5, 7, 11$$. The total number of outcomes is $$6 \times 6 = 36$$.

We count pairs $$(a, b)$$ with $$a + b \leq 8$$:

When $$a = 1$$: $$b$$ can be $$1, 2, 3, 5, 7$$ (sums: 2, 3, 4, 6, 8 — all $$\leq 8$$; $$1+11=12 > 8$$). That gives 5 outcomes.

When $$a = 2$$: $$b$$ can be $$1, 2, 3, 5$$ (sums: 3, 4, 5, 7; $$2+7=9 > 8$$). That gives 4 outcomes.

When $$a = 3$$: $$b$$ can be $$1, 2, 3, 5$$ (sums: 4, 5, 6, 8; $$3+7=10 > 8$$). That gives 4 outcomes.

When $$a = 5$$: $$b$$ can be $$1, 2, 3$$ (sums: 6, 7, 8; $$5+5=10 > 8$$). That gives 3 outcomes.

When $$a = 7$$: $$b = 1$$ only ($$7+1=8$$; $$7+2=9 > 8$$). That gives 1 outcome.

When $$a = 11$$: no valid $$b$$ ($$11+1=12 > 8$$). That gives 0 outcomes.

Total favourable = $$5 + 4 + 4 + 3 + 1 = 17$$. The probability is $$\frac{17}{36}$$.

The answer is Option B: $$\frac{17}{36}$$.