MCQ
An ordinary cube has four blank faces, one face marked $2$ another marked $3$. Then the probability of obtaining a total of exactly $12$ in $5$ throws, is
- A$\frac{5}{{1296}}$
- B$\frac{5}{{1944}}$
- ✓$\frac{5}{{2592}}$
- DNone of these
$A$ total of $12$ in $5$ throw can be obtained in following two ways -
$(i)$ One blank and four $3's = {}^5{C_1} = 5$
or $(ii)$ Three $2's$ and two $3's = {}^5{C_2} = 10$
Hence, the required probability $ = \frac{{15}}{{{6^5}}} = \frac{5}{{2592}}.$
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| Column $I$ | Column $II$ |
| $(A)$ The set $\left\{\operatorname{Re}\left(\frac{2 i z}{1-z^2}\right): z\right.$ is a complex number, $\left.|z|=1, z \neq \pm 1\right\}$ is | $(p)$ $(-\infty,-1) \cup(1, \infty)$ |
| $(B)$ The domain of the function $f(x)=\sin ^{-1}\left(\frac{8(3)^{x-2}}{1-3^{2(x-1)}}\right)$ is is | $(q)$ $(-\infty, 0) \cup(0, \infty)$ |
| $(C)$ If $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$, then the set $\left\{f(\theta): 0 \leq \theta<\frac{\pi}{2}\right\}$ is | $(r)$ $[2, \infty)$ |
| $(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$, then $f(x)$ is increasing in | $(s)$ $(-\infty,-1] \cup[1, \infty)$ |
| $(t)$ $(-\infty, 0] \cup[2, \infty)$ |