MCQ
Two dice are thrown simultaneously. The probability of getting the sum $2$ or $8$ or $12$ is
- A$\frac{5}{{18}}$
- ✓$\frac{7}{{36}}$
- C$\frac{7}{{18}}$
- D$\frac{5}{{36}}$
The sum $8$ can be found in five ways
$i.e.$ $(6,\,\,2),$ $(6,\,\,2),\,(5,\,\,3),$ $\,(4,\,\,4),\,(3,\,\,5),\,(2,\,\,6)$.
Similarly the sum twelve can be found in one way $i.e.,$ $(6,\,\,6).$
Hence required probability $ = \frac{7}{{36}}.$
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