A) \[\frac{1}{12}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{3}\]
D) \[\frac{2}{3}\]
Correct Answer: D
Solution :
Trick : As we know, the sum will be either even or odd but even is more likely to occur than odd (given). Therefore, the probability is greater than \[\frac{1}{2}\] which is given in only one option.
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