If $A$ and $B$ are two events of a random experiment such that $P(\bar{A})=\frac{2}{3}$, $P(B)=\frac{4}{15}$ and $P(A \cap \bar{B})=\frac{1}{5}$, then $\sqrt{195[P(B \mid(A \cup \bar{B}))+P(A \cup B)]} = $

Box $1$ contains three cards bearing numbers $1, 2, 3$; box $2$ contains five cards bearing numbers $1, 2, 3, 4, 5$; and box $3$ contains seven cards bearing numbers $1, 2, 3, 4, 5, 6, 7$. $A$ card is drawn from each of the boxes. Let $x_i$ be the number on the card drawn from the $i^{\text{th}}$ box,$i = 1, 2, 3$.
$1.$ The probability that $x_1 + x_2 + x_3$ is odd is:
$(A) \frac{29}{105}$ $(B) \frac{53}{105}$ $(C) \frac{57}{105}$ $(D) \frac{1}{2}$
$2.$ The probability that $x_1, x_2, x_3$ are in an arithmetic progression is:
$(A) \frac{9}{105}$ $(B) \frac{10}{105}$ $(C) \frac{11}{105}$ $(D) \frac{7}{105}$
Give the answers for question $1$ and $2$.

$A$ biased die is marked with numbers $2, 4, 8, 16, 32, 32$ on its faces. The probability of getting a face with mark $n$ is $\frac{1}{n}$. If the die is thrown thrice,then the probability that the sum of the numbers obtained is $48$ is:

Let the mean and variance of $7$ observations $2, 4, 10, x, 12, 14, y$ (where $x > y$) be $8$ and $16$ respectively. Two numbers are chosen from the set $\{1, 2, 3, x-4, y, 5\}$ one after another without replacement. Then the probability that the smaller number among the two chosen numbers is less than $4$ is:

If $10$ different balls are to be placed in $4$ distinct boxes at random,then the probability that two of these boxes contain exactly $2$ and $3$ balls is

Three numbers are chosen at random,one after another with replacement,from the set $S = \{1, 2, 3, \ldots, 100\}$. Let $p_1$ be the probability that the maximum of the chosen numbers is at least $81$ and $p_2$ be the probability that the minimum of the chosen numbers is at most $40$.
$(1)$ The value of $\frac{625}{4} p_1$ is
$(2)$ The value of $\frac{125}{4} p_2$ is