A) \[\frac{1}{27}\]
B) \[\frac{1}{9}\]
C) \[\frac{4}{27}\]
D) \[\frac{1}{6}\]
Correct Answer: D
Solution :
Total no. of ways placing 3 letters in three envelops \[=3\,!,\] out of these ways only one way is correct. Hence the required probability \[=\frac{1}{3\,!}=\frac{1}{6}.\]
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