For the three events $A, B$ and $C$,$P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs) = $P$ (exactly one of the events $C$ or $A$ occurs) = $p$ and $P$ (all the three events occur simultaneously) = $p^2$,where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is

  • A
    $\frac{3p + 2p^2}{2}$
  • B
    $\frac{p + 3p^2}{4}$
  • C
    $\frac{p + 3p^2}{2}$
  • D
    $\frac{3p + 2p^2}{4}$

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$A$ 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

Match the statements in column-$I$ with those in column-$II$.
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$(A)$ $A$ 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 $(s)$ $5$
$(t)$ $6$

$A$ bag contains $2n$ coins,out of which $n-1$ are unfair with heads on both sides and the remaining are fair. One coin is picked from the bag at random and tossed. If the probability that a head appears in the toss is $\frac{41}{56}$,then the number of unfair coins in the bag is:

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