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14.Probability
easy
The chances of throwing a total of $3$ or $5$ or $11$ with two dice is
A
$\frac{5}{{36}}$
B
$\frac{1}{9}$
C
$\frac{2}{9}$
D
$\frac{{19}}{{36}}$
Solution
(c) Total cases $ = 36$. Favourable cases $ = 2 + 4 + 2 = 8$
$\therefore $ The required probability $ = \frac{8}{{36}} = \frac{2}{9}.$
Standard 11
Mathematics