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If the mean and variance of a binomial distribution are 4 and 3, respectively, the probability of getting exactly six successes in this distribution is:
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{10}\big(\frac{3}{4}\big)^6$
$\text{ }^{16}\text{C}_6\big(\frac{1}{4}\big)^{6}\big(\frac{3}{4}\big)^{10}$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)\big(\frac{3}{4}\big)^6$
$\text{ }^{12}\text{C}_6\big(\frac{1}{20}\big)^6\big(\frac{3}{4}\big)^6$
| List $I$ | List $II$ |
| $P.$ Let $y(x)=\cos \left(3 \cos ^{-1} x\right), x \in[-1,1], x \neq \pm \frac{\sqrt{3}}{2}$. Then $\frac{1}{y(x)}\left\{\left(x^2-1\right) \frac{d^2 y(x)}{d x^2}+x \frac{d y(x)}{d x}\right\}$ equals | $1.$ $1$ |
| $Q.$ Let $A_1, A_2, \ldots \ldots, A_n(n>2)$ be the vertices of a regular polygon of $n$ sides with its centre at the origin. Let $\vec{a}_k$ be the position vector of the point $A_k, k=1,2, \ldots, n$. If $\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_k} \times \overrightarrow{a_{k+1}}\right)\right|=\left|\sum_{k=1}^{n-1}\left(\overrightarrow{a_k} \cdot \overrightarrow{a_{k+1}}\right)\right|$, then the minimum value of $n$ is | $2.$ $2$ |
| $R.$ If the normal from the point $P(h, 1)$ on the ellipse $\frac{x^2}{6}+\frac{y^2}{3}=1$ is perpendicular to the line $x+y=8$, then the value of $h$ is | $3.$ $8$ |
| $S.$ Number of positive solutions satisfying the equation $\tan ^{-1}\left(\frac{1}{2 x+1}\right)+\tan ^{-1}\left(\frac{1}{4 x+1}\right)=\tan ^{-1}\left(\frac{2}{x^2}\right)$ is | $4.$ $9$ |
Codes: $ \quad P \quad Q \quad R \quad S $