Question

$n$ letters to each of which corresponds on addressed envelopeare placed in the envelope at random. Then the probability that $n$ letter is placed in the right envelope, will be:

• A. $\frac{1}{1 !}-\frac{1}{2 !}+\frac{1}{3 !}-\frac{1}{4 !}+\ldots .(-1)^{ n } \frac{1}{ n !}$
• B. $\frac{1}{2 !}+\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \cdot \frac{1}{n !}$
• C. $\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots . .(-1)^{ n } \frac{1}{ n !}$
• D. None of these

Answer 1

Answer: (C)

Probability of $n$ balls $=1- P \left( A _{1} \cup A _{2} \cup A _{3} \cup \ldots \cup A _{ n }\right)$ Where $A_{1} \ldots . . A_{n}$ the event that the letter is placed at right envelope.
$=1-\left[\Sigma P \left( A _{ i }\right)-\Sigma P \left( A _{ i } \cap A _{ k }\right)\right.
\left.+2 P \left( A _{ i } \cap A _{ j } \cap A _{ k }\right) \ldots \ldots+(-1)^{ n -1} P \left( A _{ i } \cap A _{ j } \cap A _{ n }\right)\right]$
Here, $ P \left( A _{ i }\right)=\frac{( n -1) !}{ n !}$
$P \left( A _{1} \cap A _{2} \cap A _{3} \cap \ldots \ldots \cap A _{ n }\right)=\frac{( n - r ) !}{ n !}$
$\Rightarrow \Sigma \overline{ A }_{1} \cap \overline{ A }_{2} \cap \overline{ A }_{3} \cap \ldots \ldots \ldots \cap \overline{ A }_{ n }$
$=1-\left[\frac{1}{1 !}-\frac{1}{2 !}+\frac{1}{3 !} \ldots \ldots .(-1) \frac{ n -1 !}{ n !}\right]$
$=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\ldots \ldots \ldots+(-1)^{ n } \frac{1}{ n !}$