- 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 $p$ : The value of $sin\,120^o$ can be divided by taking $\theta\, = 240^o$ in the equation $2\,\sin \frac{\theta }{2} = \sqrt {1 + \sin \theta } - \sqrt {1 - \sin \theta } $
Statement $q$ : The angles $A, B, C$ and $D$ of any quadrilateral $ABCD$ satisfy the equation $\cos \left( {\frac{1}{2}\left( {A + C} \right)} \right) + \cos \left( {\frac{1}{2}\left( {B + D} \right)} \right) = 0$
Then the truth values of $p$ and $q$ are respectively.