$A$ person goes to office by car, scooter, bus, and train, the probabilities of which are $1/7, 3/7, 2/7,$ and $1/7$ respectively. The probability that he reaches office late if he takes a car, scooter, bus, or train is $2/9, 1/9, 4/9,$ and $1/9$ respectively. Given that he reached the office in time, the probability that he travelled by car is: (in $/7$)

In four schools $B_1, B_2, B_3, B_4$,the percentage of girl students is $12, 20, 13, 17$ respectively. From a school selected at random,one student is picked up at random and it is found that the student is a girl. The probability that the school selected is $B_2$ is:

$A$ box $B_1$ contains $3$ blue balls and $6$ red balls. Another box $B_2$ contains $8$ blue balls and $n$ red balls $(n \in N)$. $A$ ball selected at random from a box is found to be red. If $p$ is the probability that this red ball drawn is from box $B_2$,then

$A$ card from a pack of $52$ cards is lost. From the remaining $51$ cards,$n$ cards are drawn and are found to be spades. If the probability of the lost card being a spade is $\frac{11}{50}$,then $n$ is equal to

Two balls are selected at random one by one without replacement from a bag containing $4$ white and $6$ black balls. If the probability that the first selected ball is black,given that the second selected ball is also black,is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $m + n$ is equal to :

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 $2$.