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, $P\left(A_{i}\right)=\frac{\left(n-1\right)!}{n!}$
$P\left(A_{1}\cap A_{2}\cap A_{3}\cap......\cap A_{n}\right)=\frac{\left(n-r\right)!}{n!}$
$\Rightarrow \sum\bar{A}\cap\bar{A}_{2}\cap\bar{A}_{3}\cap........\cap\bar{A}_{n}$
$=1-\left[\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}......\left(-1\right) \frac{n-1!}{n!}\right]$
$=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-........+\left(-1\right)^{n} \frac{1}{n!}$