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14.Probability
easy
Probability of throwing $16$ in one throw with three dice is
A
$\frac{1}{{36}}$
B
$\frac{1}{{18}}$
C
$\frac{1}{{72}}$
D
$\frac{1}{9}$
Solution
(a) Total number of favorable ways
$n\,(E) = \{ (6,6,4)\} ,\,(6,4,6),\,(4,6,6),\,(5,5,6),\,(5,6,5)(6,5,5)\} = 6$
Total number of ways $= n(S) = 6 \times 6 \times 6 = 216$
$\therefore $Required probability $= \frac{{n\,(E)}}{{n\,(S)}} = \frac{6}{{216}} = \frac{1}{{36}}$.
Standard 11
Mathematics
