MCQ
The chance of getting a doublet with $2$ dice is
- A$\frac{2}{3}$
- ✓$\frac{1}{6}$
- C$\frac{5}{6}$
- D$\frac{5}{{36}}$
Favourable number of outcomes $= 6$
$i.e.$, $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6),$
$\therefore$ Required probability $ = \frac{6}{{36}} = \frac{1}{6}.$
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