The urns $A, B$ and $C$ contain $4$ red,$6$ black; $5$ red,$5$ black and $\lambda$ red,$4$ black balls respectively. One of the urns is selected at random and a ball is drawn. If the ball drawn is red and the probability that it is drawn from urn $C$ is $0.4$,then find the square of the length of the side of the largest equilateral triangle inscribed in the parabola $y^2 = \lambda x$ with one vertex at the vertex of the parabola.

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 $2$.

$A$ computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20 \%$ and plant $T_2$ produces $80 \%$ of the total computers produced. $7 \%$ of computers produced in the factory turn out to be defective. It is known that $P(\text{defective} | T_1) = 10 P(\text{defective} | T_2)$. $A$ computer produced in the factory is randomly selected and it is not defective. Then the probability that it is produced in plant $T_2$ is

$A$ box contains $n$ coins,$m$ of which are fair and the rest are biased. When a biased coin is tossed,the probability of getting a head is twice as likely as tail. $A$ coin is drawn from the box at random and is tossed twice. It is found that the first time it shows head and the second time it shows tail. Then,the probability that the coin drawn is fair is

Suppose we have four boxes $A, B, C$ and $D$ containing coloured marbles as given below:

Box	Red	White	Black
$A$	$1$	$6$	$3$
$B$	$6$	$2$	$2$
$C$	$8$	$1$	$1$
$D$	$0$	$6$	$4$

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red,what is the probability that it was drawn from box $A$,box $B$,or box $C$?

$A$ bag contains $19$ unbiased coins and one coin with head on both sides. One coin is drawn at random and tossed,and a head turns up. If the probability that the drawn coin was unbiased is $\frac{m}{n}$,where $\operatorname{gcd}(m, n) = 1$,then $n^2 - m^2$ is equal to: