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Q. Two fair dice are rolled. The probability of the sum of digits on their faces to be greater than or equal to $10$ is

EAMCETEAMCET 2013

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
$=\frac{n(E)}{n(S)}=\frac{6}{36}=\frac{1}{6}$