$A$ dealer gets refrigerators from $3$ different manufacturing companies $C_1, C_2$ and $C_3$. $25 \%$ of his stock is from $C_1, 35 \%$ from $C_2$ and $40 \%$ from $C_3$. The percentages of receiving defective refrigerators from $C_1, C_2$ and $C_3$ are $3 \%, 2 \%$ and $1 \%$ respectively. If a refrigerator sold at random is found to be defective by a customer, then the probability that it is from $C_2$ is

Three companies $C_1, C_2, C_3$ produce car tyres. $A$ car manufacturing company buys $40 \%$ of its requirement from $C_1, 35 \%$ from $C_2$ and $25 \%$ from $C_3$. The company knows that $2 \%$ of the tyres supplied by $C_1, 3 \%$ by $C_2$ and $4 \%$ by $C_3$ are defective. If a tyre chosen at random from the consignment received is found defective, then the probability that it was supplied by $C_2$ 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

Bag $I$ contains $3$ red,$4$ black,and $3$ white balls. Bag $II$ contains $2$ red,$5$ black,and $2$ white balls. One ball is transferred from Bag $I$ to Bag $II$,and then a ball is drawn from Bag $II$. The ball drawn is found to be black. What is the probability that the transferred ball was red?

$A$ box $A$ contains $2$ white,$3$ red and $2$ black balls. Another box $B$ contains $4$ white,$2$ red and $3$ black balls. If two balls are drawn at random,without replacement,from a randomly selected box and one ball turns out to be white while the other ball turns out to be red,then the probability that both balls are drawn from box $B$ 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 $2$.