Three students S_1, S_2 and S_3 are given a p | vedclass

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(A) $P(U) = 1 - P(S_1^{\prime} \cap S_2^{\prime} \cap S_3^{\prime}) = \frac{1}{2}$$\Rightarrow P(S_1^{\prime}) \cdot P(S_2^{\prime}) \cdot P(S_3^{\pri

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$\frac{13}{36}$

Vedclass · Answer

(A) $P(U) = 1 - P(S_1^{\prime} \cap S_2^{\prime} \cap S_3^{\prime}) = \frac{1}{2}$$\Rightarrow P(S_1^{\prime}) \cdot P(S_2^{\prime}) \cdot P(S_3^{\prime}) = \frac{1}{2}$$\Rightarrow (1 - P(S_1))(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2} \dots (1)$$P(V) = \frac{P(S_1 \cap S_2^{\prime} \cap S_3^{\prime})}{P(S_2^{\prime} \cap S_3^{\prime})} = \frac{1}{10}$$\Rightarrow \frac{P(S_1) \cdot P(S_2^{\prime}) \cdot P(S_3^{\prime})}{P(S_2^{\prime}) \cdot P(S_3^{\prime})} = \frac{1}{10}$$\Rightarrow P(S_1) = \frac{1}{10}$$P(W) = P(S_2 \cap S_3^{\prime}) = P(S_2) \cdot P(S_3^{\prime}) = \frac{1}{12}$$\Rightarrow P(S_2)(1 - P(S_3)) = \frac{1}{12} \dots (2)$From Eq. $(1)$,$(1 - \frac{1}{10})(1 - P(S_2))(1 - P(S_3)) = \frac{1}{2}$$\Rightarrow (1 - P(S_2))(1 - P(S_3)) = \frac{5}{9} \dots (3)$Dividing Eq. $(2)$ by Eq. $(3)$: $\frac{P(S_2)}{1 - P(S_2)} = \frac{1}{12} 	imes \frac{9}{5} = \frac{3}{20}$$20 P(S_2) = 3 - 3 P(S_2) \Rightarrow 23 P(S_2) = 3 \Rightarrow P(S_2) = \frac{3}{23}$Substituting $P(S_2)$ in Eq. $(2)$: $\frac{3}{23}(1 - P(S_3)) = \frac{1}{12}$$1 - P(S_3) = \frac{23}{36} \Rightarrow P(S_3) = 1 - \frac{23}{36} = \frac{13}{36}$Thus,$P(T) = P(S_3) = \frac{13}{36}$.

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

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