Solving for N in a binomial distribution Mid-term study...

Two dice are rolled. How many times must the dice be rolled so that the probability of getting a sum of 10 or greater on at least one roll is larger than 0.9?

So am I correct in thinking that it would be:
$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k}P^k(1-P)^{j-k} \geq .9$$
Given that the probability of getting a sum of 10+ is $\frac{6}{36}$
I get:
$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k} \left( \frac{6}{36}\right)^k \left( \frac{30}{36}\right)^{j-k} \geq .9$$
Am I right in my formula? If not, what should it be? And then how do I solve for N?
Thanks.

EDIT / RESPONSE:
What I wrote above is the same as saying
$${n \choose 0}P^0(1-P)^n < .1$$
The first parts are both just 1, which plugging in numbers leaves me with
$$\left(\frac{30}{36} \right)^n < .1$$
To solve, I just need to do the following
$$\log_{\frac{30}{36}}{\left(\frac{30}{36} \right)^n} < log_{\frac{30}{36}}{(.1)}$$
Which according to my calculations should be 12.6, which means 13.


So am I correct in thinking that it would be
$$\sum_{j=1}^{n}\sum_{k=1}^{j}{n \choose k}P^k(1-P)^{j-k} \geq .9$$

No, not quite.  
Rolling two die such that their sum is less than 10 is a trial with probability of success $p$.  
The number of successes, $N_n$, in a series of $n$ trials has a binomial probability distribution.  $N_n\sim \mathcal{Bin}(n, p)$.
You simply want to find the number of trials, $n$, such that $\mathsf P(N_n\geq 1)\geq 0.9$.  That is:
$$ \sum_{k=1}^n {n\choose k}p^k(1-p)^{n-k}\geq 0.9$$
However the probability of the complement event is easier to much calculate, so find $n$ such that: $\mathsf P(N_n=0) \leq 0.1$.  That is:
$$(1-p)^n \leq 0.1$$
