Find the least number of balls that must be taken from the bag.. I'm building onto my previous question, but I will include it also in here. I'm again struggling with the probability questions we've been given.
Six balls numbered 1,2,2,3,3,3 are placed in a bag. Balls are taken one at a time from the bag at random and the number noted. Throughout the question a ball is always replaced before the next ball is taken.
The questions are following:
A) Find the least number of balls that must be taken from the bag for the probability of taking out at least one ball numbered 2 to be greater than 0.95.
B) Another bag also contains balls numbered 1,2 or 3. Eight balls are taken from the bad at random. It is calculated that the expected number of balls numbered 1 is 4.8, and the variance of the number of balls numbered 2 is 1.5. Find the least possible number of balls numbered 3 in this bag. 
I would really appreciate your help or at least hints..

a) Let $X_i$ denote the label of the $i^{th}$ drawn ball and let $N=\inf\{i>0:X_i=2\}$. Then $N\sim Geo(2/6)$, since there are $6$ balls and $2$ of them are numbered $2$. 
$$ \mathbb P(N=n) = \bigg(1-\frac{1}{3} \bigg)^{n-1}\frac13 \quad n\geq 1. $$
Let $A_n$ denote the event that among $n$ draws there was at least 1 ball numbered 2, then
$$ \mathbb P(A_n) = \mathbb P(N\leq n) = \sum_{k=1}^n(\frac{2}{3})^{k-1} \frac{1}{3} = 1-(\frac23)^n \geq 0.95  $$
whence
$$ n  \geq \frac{\log 0.05}{\log 2/3}. $$
b) Let $X_1$ denote the number of balls numbered $1$ from the 8 drawn balls. Then $X_1\sim Binom(8,p_1)$ and $\mathbb E X_1 = 8p_1 = 4.8 \Rightarrow p_1 = 0.6 $. Also $X_2\sim Binom(8,p_2)$ and $\mathbb D^2(X_2) = 8p_2(1-p_2) = 1.5$
$$ 8p_2^2-8p_2+1.5 = 0 $$
$$ p_2 = \frac{8\pm \sqrt{64-4\cdot8\cdot 1.5}}{16} = \begin{cases} 3/4 & \mbox{or} \\ 1/4.
\end{cases} $$
Let $p_3$ denote the probability that a ball numbered 3 is drawn. Then $p_1+p_2+p_3=1$ has to hold, thus $p_2=1/4$, and so $p_3 =0.15 $. Let $n$ denote the total number of balls in the bag. Then $p_i = n_i / n$, hence $ \mathbb N\ni n_i = np_i $ for all $i$.  From this one gets that $n\geq 20$ and $n_3 \geq 3$
