A) \[\frac{1}{2}\]
B) \[\frac{1}{6}\]
C) \[\frac{2}{3}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
Total number of possible outcomes \[=6\times 6=36\] Favourable outcomes = (1, 1), (2, 2), (3, 3) (4, 4), (5, 5) and (6, 6). Total number of favourable outcomes = 6 \[P\left( E \right)=\frac{6}{36}=\frac{1}{6}\]
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