Probability that at least two $9$'s appear as sum in four tosses of a Pair of fair Dice.

Question

I can calculate in the following way :

1. $P(9 \text{ in one toss}) = 1/9$
2. sum the following : $P(9 \text{ in two toss})$, $P(9 \text{ in three toss})$, $P(9 \text{ in 4 toss})$

And the answer I get is $417/6561$.

But is there any other/better way to solve this ? (new to probability , sorry if this is a stupid question)

Answer 1

Hint: Better if use $P(x \geq 2) = 1 -P(x=0) - P(x=1)$