A) \[\frac{1}{3}\]
B) \[\frac{1}{6}\]
C) \[\frac{5}{12}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
| [b] Total number of outcomes = 36 |
| Let E be the event 'doublet.' |
| So, outcomes favourable to E are |
| \[\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}\]i.e.. 6 in number. |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,P(E)=\frac{6}{36}=\frac{1}{6}\] |
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