Let $n_1$ and $n_2$ be the number of red and black balls, respectively, in box $I$. Let $n_3$ and $n_4$ be the number of red and black balls, respectively, in box $II$.
$1.$ One of the two boxes, box $I$ and box $II$, was selected at random and a ball was drawn randomly out of this box. The ball was found to be red. If the probability that this red ball was drawn from box $II$ is $\frac{1}{3}$, then the correct option$(s)$ with the possible values of $n_1, n_2, n_3$ and $n_4$ is(are)
$(A)$ $n_1=3, n_2=3, n_3=5, n_4=15$
$(B)$ $n_1=3, n_2=6, n_3=10, n_4=50$
$(C)$ $n_1=8, n_2=6, n_3=5, n_4=20$
$(D)$ $n_1=6, n_2=12, n_3=5, n_4=20$
$2.$ A ball is drawn at random from box $I$ and transferred to box $II$. If the probability of drawing a red ball from box $I$, after this transfer, is $\frac{1}{3}$, then the correct option$(s)$ with the possible values of $n_1$ and $n_2$ is(are)
$(A)$ $n_1=4, n_2=6$ $(B)$ $n_1=2, n_2=3$
$(C)$ $n_1=10, n_2=20$ $(D)$ $n_1=3, n_2=6$
Give the answer question $1$ and $2.$