A) \[\frac{5}{36}\]
B) \[\frac{11}{36}\]
C) \[\frac{1}{6}\]
D) \[\frac{1}{3}\]
Correct Answer: B
Solution :
Favourable cases for one are three i.e. 2, 4 and 6 and for other are two i.e. 3, 6. Hence required probability \[=\left[ \left( \frac{3\times 2}{36} \right)\text{ }2-\frac{1}{36} \right]=\frac{11}{36}\] {As same way happen when dice changes numbers among themselves}
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