A computer producing factory has only two plants $T_1$ and $T_2$. Plant $T_1$ produces $20\%$ and plant $T_2$ produces $80\%$ of the total computers produced. $7\%$ of computers produced in the factory turn out to be defective. It is known that,  $P$(computer turns out to be defective given that it is produced in plant $T_1$) $=10P$(computer turns out to be defective given that it is produced in plant $T_2$). Where, $P\left(E\right)$ denotes the probability of an event $E.$ A computer produced in the factory is randomly selected and it does not turn out to be defective. Then the probability that it is produced in plant $T_2$ is

Let $n_1 and n_2$ be the number of red and black balls, respectively, in box I. Let $n_{3}and n_4$ be the number of red and black balls, respectively, in box II.
One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of this box. The balls was found to be red. If the probability that this red ball was drawn from box II is $\frac{1}{3}$ , then the correct option(s) with the possible values of $n_1, n_2, n_{3}and n_4$ is(are)

For \(0 < \theta < \frac{\pi}{2}\), the solution(s) of \(\sum_{m=1}^6 \operatorname{cosec}\left[\theta+\frac{(m-1) \pi}{4}\right]\) \(\operatorname{cosec}\left(\theta+\frac{m \pi}{4}\right)=4 \sqrt{2}\) is/are

Let \(M\) and \(N\) be two \(3 \times 3\) non-singular skew-symmetric matrices such that \(M N=N M\). If \(P^T\) denotes the transpose of \(P\), then \(M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T\) is equal to

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Read the following passage and answer the questions.
Let \(A B C D\) be a square of side length 2 units. \(C_2\) is the circle through vertices \(A, B, C, D\) and \(C_1\) is the circle touching all the sides of square \(A B C D\). \(L\) is the line through \(A\).Question:
If \(P\) is a point on \(C_1\) and \(Q\) is a point on \(C_2\), then \(\frac{P A^2+P B^2+P C^2+P D^2}{Q A^2+Q B^2+Q C^2+Q D^2}\) is equal to

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Let \(V_r\) denotes the sum of the first \(r\) terms of an arithmetic progression \((A P)\) whose first term is \(r\) and the common difference is \((2 r-1)\). Let \(T_r=V_{r+1}-V_r-2\) and \(Q_r=T_{r+1}-T_r\) for \(r=1,2, \ldots\)
Question:
The sum \(V_1+V_2+\ldots+V_n\) is

Three students \(S_1, S_2\) and \(S_1\) are given a problem to solve. Consider the following events :
\(U:\) At least one of \(S_1, S_2\) and \(S_3\) can solve the problem,
\(V: S_1\) can solve the problem, given that neither \(\mathrm{S}_2\) nor \(\mathrm{S}_3\) can solve the problem,
\(W: S_2\) can solve the problem and \(S_3\) cannot solve the problem,
\(T: S_3\) can solve the problem.
For any event \(E\), let \(P(E)\) denote the probability of \(E\).
If \(P(U)=\frac{1}{2}, P(V)=\frac{1}{10}\) and \(P(W)=\frac{1}{12}\), then \(P(T)\) is equal to

Based on VSEPR model, match the xenon compounds given in List-I with the corresponding geometries and the number of lone pairs on xenon given in List-II and choose the correct option.

List - I	List - II
(P) \(XeF _2\)	(1) Trigonal bipyramidal and two lone pair of electrons
(Q) \(XeF _4\)	(2) Tetrahedral and one lone pair of electrons
(R) \(XeO _3\)	(3) Octahedral and two lone pair of electrons
(S) \(XeO _3 F_2\)	(4) Trigonal bipyramidal and no lone pair of electrons
	(5) Trigonal bipyramidal and three lone pair of electrons

The following integral $\underset{\frac{\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}{\left(2\mathrm{ }\mathrm{c}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{c}\mathrm{ }\mathrm{x}\right)}^{17}\mathrm{ }\mathrm{d}\mathrm{x}$ is equal to

A composite block is made of slabs \(A, B, C, D\) and \(E\) of different thermal conductivities (given in terms of a constant, \(K\) ) and sizes (given in terms of length, \(L\) ) as shown in the figure. All slabs are of same width. Heat \(Q\) flows only from left to right through the blocks. Then, in steady state

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Tollen's reagent is used for the detection of aldehyde when a solution of \(\mathrm{AgNO}_3\) is added to glucose with \(\mathrm{NH}_4 \mathrm{OH}\) then gluconic acid is formed
\(\mathrm{Ag}^{+}+e^{-} \rightarrow \mathrm{Ag} E_{\text {red }}^{\circ}=0.8 \mathrm{~V} \)
\( \mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6+\mathrm{H}_2 \mathrm{O} \rightarrow \text { Gluconic acid }\)\(\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_7\right)+2 \mathrm{H}^{+}+2 e^{-} ; -E_{\mathrm{red}}^{\circ}=\) \(-0.05 \mathrm{~V} \)
\( \mathrm{Ag}\left(\mathrm{NH}_3\right)_2^{+}+e^{-} \rightarrow \mathrm{Ag}(s)+2 \mathrm{NH}_3 ;\) \(E_{\mathrm{red}}^{\circ}=0.337 \mathrm{~V}\)
[Use \(2.303 \times \frac{R T}{F}=0.0592\) and \(\frac{F}{R T}=38.92\) at \(298 \mathrm{~K}\) ]
Question:
Ammonia is always added in this reaction. Which of the following must be incorrect?

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A fair die is tossed repeatedly until a six is obtained. Let \(X\) denotes the number of tosses required.Question:
The probability that \(X=3\) equals

Let $f\left(x\right)=7{\tan }^{8}x+7{\tan }^{6}x-3{\tan }^{4}x-3{\tan }^{2}x$ for all $x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right).$ Then the correct expression(S) is(are)