Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to…
Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to 10 is
- $\frac{1}{5}$
- $\frac{1}{4}$
- $\frac{1}{8}$
- $\frac{1}{6}$
Solution
Total samle points, $n(S)=6 \times 6=36$
Favourable events
$
=[(6,4),(6,5),(6,6),(5,5),(5,6),(4,6)]
$
Total favourable events, $n(E)=6$
Required probability
$
=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}
$
Asked in: AP EAMCET 2013
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