If $X=(X_1,\ldots,X_r)$ has a multinomial distribution, then each of the components $X_1,\ldots,X_r$ has a binomial distribution.
You're distributing $n$ objects into $r$ bins. For each object, the probability that it falls into the $k$th bin is $p_k,$ for $k=1,\ldots,r.$ The number of objects that fall into the $k$th bin is $X_k,$ for $k=1,\ldots,r.$ So $X_1+X_2$ is the number of objects falling into either of the first two bins. The probability that an object falls into either of the first two bins is the sum of the probabilities of its falling into those two bins, i.e. it is $p_1+p_2.$ In effect, you've simply joined those two bins together, so now you have $r-1$ bins, with probabilities $p_1+p_1,p_3,p_4,\ldots,p_r$ of an object falling into them. Therefore the distribution of $(X_1+X_2,X_3,X_4,\ldots,X_r)$ is multinomial with parameters $(n,p_1+p_2, p_3, p_4, \ldots, p_r).$ And as before, each component separately has a binomial distribution.
Best Answer
$\Pr(X \ge 1)=1-\Pr(X=0)=1-\left(\frac23\right)^n$.
If this is greater than or equal to $0.85$ then $\left(\frac23\right)^n \le 0.15$, i.e. $n \log(2/3) \le \log(0.15)$.
Note the negative logarithms and you get $n \ge \dfrac{\log(20)-\log(3)}{\log(3)-\log(2)}$.
Calculate and round up to an integer.