A) \[\frac{1}{6}\]
B) \[\frac{1}{18}\]
C) \[\frac{2}{9}\]
D) \[\frac{23}{108}\]
Correct Answer: C
Solution :
\[n(S)=36\] E = Event of getting sum 7 ie, \[\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}\] \[\therefore \] \[n(F)=2\] \[F=\]Event of getting sum 11 ie, \[\{(6,5),(5,6)\}\] \[\therefore \] \[n(F)=2\] Also, \[n(E\cap F)=0\] \[\therefore \] \[n(E\cup F)=n(E)+n(F)-n(E\cap F)\] \[=6+2=8\] \[\therefore \]Required probability\[=\frac{8}{36}=\frac{2}{9}\] Alternative Method Probability of getting sum of 7 is\[\frac{6}{36}\]and Probability of getting sum of 11 is\[\frac{2}{36}\]. \[\therefore \]Probability of getting sum 7 or 11 is \[\frac{6}{36}+\frac{2}{36}=\frac{8}{36}=\frac{2}{9}\]
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