Question

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

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

Answer 1

Answer: (C)

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