A) \[\frac{2}{3}\]
B) \[\frac{1}{3}\]
C) \[\frac{1}{9}\]
D) \[\frac{7}{36}\]
Correct Answer: B
Solution :
| [c] Total number of possible outcomes |
| \[=6\times 6=36\] |
| Favourable outcomes |
| = The produce of the two numbers on the top of the dice |
| \[=\{(3,2),(2,3),(6,1),(1,6)\}\] |
| \[\therefore \] Total number of favourable outcomes = 4 |
| So, the required probability \[=\frac{4}{36}=\frac{1}{9}\]. |
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