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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Statement $1:$ If arg $Z+$ arg $W = \pi ,$ then $Z = -\overline W $.
Statement $2:$ $\left| Z \right| = \left| W \right|,$ implies arg $Z-$ arg $\overline W = \pi .$