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(B) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B)$.This implies $\frac{n(A \cap B)}{n(S)} = \frac{n(A)}{n(S)} 	imes \frac{n(B)}{n(

Vedclass · Accepted Answer

$422$

Vedclass · Answer

(B) Since $A$ and $B$ are independent events,$P(A \cap B) = P(A)P(B)$.This implies $\frac{n(A \cap B)}{n(S)} = \frac{n(A)}{n(S)} 	imes \frac{n(B)}{n(S)}$,so $n(A \cap B) = \frac{n(A)n(B)}{6}$.Since $n(A \cap B)$ must be an integer,$n(A)n(B)$ must be a multiple of $6$.Given $1 \leq |B| $1$. If $n(A) = 3, n(B) = 2$,then $n(A \cap B) = \frac{3 	imes 2}{6} = 1$. Number of pairs $= \binom{6}{3} 	imes \binom{3}{1} 	imes \binom{3}{1} = 20 	imes 3 	imes 3 = 180$.$2$. If $n(A) = 4, n(B) = 3$,then $n(A \cap B) = \frac{4 	imes 3}{6} = 2$. Number of pairs $= \binom{6}{4} 	imes \binom{4}{2} 	imes \binom{2}{1} = 15 	imes 6 	imes 2 = 180$.$3$. If $n(A) = 6$,then $n(B)$ can be $1, 2, 3, 4, 5$. For $n(A)=6$,$P(A)=1$,so $P(A \cap B) = P(B)$,which is always true for any $B \subseteq S$. The number of such subsets $B$ is $2^6 = 64$. Excluding $B = \emptyset$ $(|B|=0)$,we have $64 - 1 = 63$. However,we must satisfy $|B| Summing up: $180 + 180 + 62 = 422$ (adjusting for specific constraints).

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	$(s)$ $5$
	$(t)$ $6$

Let $|X|$ denote the number of elements in set $X$. Let $S = \{1, 2, 3, 4, 5, 6\}$ be a sample space,where each element is equally likely to occur. If $A$ and $B$ are independent events associated with $S$,then the number of ordered pairs $(A, B)$ such that $1 \leq |B| < |A|$ equals:

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