A) \[\frac{2}{3}\]
B) \[\frac{1}{6}\]
C) \[\frac{1}{3}\]
D) \[\frac{5}{6}\]
Correct Answer: D
Solution :
| [d] Total number of possible outcomes = 36 |
| Let E be the event getting same faces. |
| Favourable outcomes to event E are |
| \[\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}\]i.e., 6 in number, |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,p(E)=\frac{6}{36}=\frac{1}{6}\] |
| \[\therefore \,\,\,\,\,\,\,\,\,\,\,\,p(E)=\frac{6}{36}=\frac{1}{6}\] |
| \[\therefore \] P(getting different faces) |
| \[=P(\bar{E})=1-P(E)=1-\frac{1}{6}=\frac{5}{6}\] |
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