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14.Probability
easy
In a throw of three dice, the probability that at least one die shows up $1$, is
A
$\frac{5}{6}$
B
$\frac{{91}}{{216}}$
C
$\frac{1}{{36}}$
D
$\frac{{125}}{{216}}$
Solution
(b) Required probability is $1 – p$ (no die show up $1$)
$ = 1 – {\left( {\frac{5}{6}} \right)^3} = \frac{{216 – 125}}{{216}} = \frac{{91}}{{216}}.$
Standard 11
Mathematics
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