Two coins $A$ and $B$ are kept in an urn. When coin $A$ is flipped,the probability of getting a head is $1/4$,while for coin $B$ it is $3/4$. One coin is randomly chosen from this bag,tossed twice,and it falls heads on both occasions. The probability that it is coin $A$ is:

There are three families $F_1, F_2, F_3$. $F_1$ has $2$ boys and $1$ girl; $F_2$ has $1$ boy and $2$ girls; $F_3$ has $1$ boy and $1$ girl. $A$ family is randomly chosen and a child is chosen from that family randomly. If it is known that the child thus selected is a girl,then the probability that she is from $F_2$ is

$A$ card from a pack of $52$ cards is lost. From the remaining cards of the pack,two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond.

Bag $A$ contains $2$ white and $3$ red balls and bag $B$ contains $4$ white and $5$ red balls. If one ball is drawn at random from one of the bags and is found to be red,then the probability that it was drawn from the bag $B$ is

Box $I$ contains $30$ cards numbered $1$ to $30$ and Box $II$ contains $20$ cards numbered $31$ to $50$. $A$ box is selected at random and a card is drawn from it. The number on the card is found to be a non-prime number. The probability that the card was drawn from Box $I$ is

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 for question $1$ and $2$.