Three screws are drawn at random from a lot of $50$ screws containing $5$ defective ones. The probability of the event that all $3$ screws drawn are non-defective,assuming that the drawing is $(a)$ with replacement and $(b)$ without replacement,is respectively:

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 $2$.

For independent events $A$ and $B$,if $P(A) = \frac{1}{2}$ and $P(A \cup B) = \frac{3}{5}$,then $P(B) =$ . . . . . . .

$A$ speaks truth in $75 \%$ of the cases and $B$ in $80 \%$ of the cases. Then,the probability that their statements about an incident do not match,is

$A$ fair die is thrown until $2$ appears. Then the probability that $2$ appears in an even number of throws is

If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P(F) < 1,$ then