$A$ and $B$ throw a die alternatively till one of them gets a '$6$' and wins the game. Find their respective probabilities of winning,if $A$ starts first.

$A$ computer program has two modules $X$ and $Y$ and errors in them occur independently. $X$ has an error with probability $0.1$ and $Y$ has an error with probability $0.3$. If an error in $X$ alone causes the program to crash with probability $0.5$, an error in $Y$ alone causes the program to crash with probability $0.7$, and an error in both $X$ and $Y$ causes the program to crash with probability $0.8$, then the probability that the program crashes is

$A$ six-faced die is biased such that it is twice as likely to show an even number as an odd number when thrown. If the die is thrown twice,what is the probability that the sum of the two numbers obtained is even?

Match the statements in column-$I$ with those in column-$II$.

column-$I$	column-$II$
$(A)$ $A$ line from the origin meets the lines $\frac{x-2}{1}=\frac{y-1}{-2}=\frac{z+1}{1}$ and $\frac{x-\frac{8}{3}}{2}=\frac{y+3}{-1}=\frac{z-1}{1}$ at $P$ and $Q$ respectively. If length $PQ=d$,then $d^2$ is	$(p)$ $-4$
$(B)$ The values of $x$ satisfying $\tan ^{-1}(x+3)-\tan ^{-1}(x-3)=\sin ^{-1}\left(\frac{3}{5}\right)$ are	$(q)$ $0$
$(C)$ Non-zero vectors $\vec{a}, \vec{b}$ and $\vec{c}$ satisfy $\vec{a} \cdot \vec{b}=0$,$(\vec{b}-\vec{a}) \cdot(\vec{b}+\vec{c})=0$ and $2|\vec{b}+\vec{c}|=|\vec{b}-\vec{a}|$. If $\vec{a}=\mu \vec{b}+4 \vec{c}$,then the possible values of $\mu$ are	$(r)$ $4$
$(D)$ Let $f$ be the function on $[-\pi, \pi]$ given by $f(0)=9$ and $f(x)=\frac{\sin \left(\frac{9 x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$ for $x \neq 0$. The value of $\frac{2}{\pi} \int_{-\pi}^\pi f(x) dx$ is	$(s)$ $5$
	$(t)$ $6$

If the probabilities of three independent events occurring are $p_1, p_2, p_3$,what is the probability that at least one of these events occurs?

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happen is $\frac{1}{12}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then