A) \[\frac{1}{4}\]
B) \[\frac{5}{18}\]
C) \[\frac{7}{36}\]
D) \[\frac{1}{6}\]
Correct Answer: A
Solution :
| [a] Total number of possible outcomes \[=6\times 6=36\] |
| Favourable outcomes = Even numbers on both dice |
| \[=\{(2,4),(2,6),(4,2),(4,6),(6,2)(6,4),(2,2),(4,4),(6,6)\}\] \[\therefore \] Total number of favourable outcomes =9. |
| So, the required probability \[=\frac{9}{36}=\frac{1}{4}\] |
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