Gujarati
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

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