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

Answer 1

Answer: (C)

Probability of $n$ balls $=1-P\left(A_1\cup A_2\cup A_3\cup \dots \cup A_n\right)$ Where $A_1\dots ..A_n$ the event that the letter is placed at right envelope.
$=1-[\Sigma P\left(A_i\right)-\Sigma P\left(A_i\cap A_k\right)<br/><br/>+2P\left(A_i\cap A_j\cap A_k\right)\dots \dots +(-1{)}^{n-1}P\left(A_i\cap A_j\cap A_n\right)]$
Here, $P\left(A_i\right)=\frac{(n-1)!}{n!}$
$P\left(A_1\cap A_2\cap A_3\cap \dots \dots \cap A_n\right)=\frac{(n-r)!}{n!}$
$\Rightarrow \Sigma {\overline{A}}_1\cap {\overline{A}}_2\cap {\overline{A}}_3\cap \dots \dots \dots \cap {\overline{A}}_n$
$=1-\left[\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}\dots \dots .(-1)\frac{n-1!}{n!}\right]$
$=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\dots \dots \dots +(-1{)}^n\frac{1}{n!}$