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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