A) \[\frac{1}{n!}\]
B) \[1-\frac{1}{n!}\]
C) \[1-\frac{1}{n}\]
D) None of these
Correct Answer: B
Solution :
[b] Probability of all the letters kept in the right envelope is \[\frac{1}{n!}\](\[\because \]Total letters = n) i.e., \[P=\frac{1}{n!}\] We know, if q is the term used for the probability of the letters which are not kept in the right envelope. Then \[p+q=1\Rightarrow q=1-p=1-\frac{1}{n!}.\]
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