A) \[6/108\]
B) \[5/27\]
C) \[1/24\]
D) \[103/108\]
Correct Answer: D
Solution :
Total number of cases \[={{6}^{3}}=216\] Let A denote the event of getting at least \[6,\bar{A}\] will denote the event of getting less than 6 for three dice \[\therefore \] \[\bar{A}=\{(1,1,1),(1,1,2),(1,1,3),(1,2,1),\] \[(1,3,1),(2,1,2),(1,2,2),(2,2,1),\] \[(2,1,1),(3,1,1)\}\] \[n(\bar{A})=10\] Probability of getting less than \[6=\frac{10}{216}=\frac{5}{108}\] Now, \[P(A)=1-P(\bar{A})=1-\frac{5}{108}=\frac{103}{108}\]
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