Question

$n$ letters to each of which corresponds an addressed envelope are placed in the envelopes at random. What is the probability that no letter is placed in the right envelope?

• A. $1-\{\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\cdots +{(-1)}^n.\frac{1}{n!}\}$
• B. $\{\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\cdots +{(-1)}^n.\frac{1}{n!}\}$
• C. $\{\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}\}$
• D. $1-\{\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots +\frac{1}{n!}\}$

Answer 1

Answer: (A)

$1-\{\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\cdots +{(-1)}^n.\frac{1}{n!}\}$

Let $A_{th}$ denote the event that the $i_{th}$ letter is placed in the right envelope.
Then the required probability is $\frac{P({\overline{A}}_1\cap {\overline{A}}_2\cap .....\cap {\overline{A}}_n)}{P(A_i\cup A_2\cup ......A_n)}$.
[By De-Morgan law]
$=1-P(A_1\cup A_2\cup ....\cup A_n)$
$=1-[\sum P\left(A_i\right)-\sum p(A_i\cap A_j)]+\sum P(A_i\cap a_j\cap A_k)i\ne ji\ne j\ne k$
$-\cdots +{(-1)}^{n-1}p(A_i\cap A_2\cap \cdots \cap A_k)$
Now $p\left(A_i\right)=\frac{(n-1)!}{n!}$ as having placed $i^{th}$ letter in the right envelope, the remaining letters can be placed in $(n-1)!$ ways.
Similarly, $P(A_1\cap A_2\cap .....\cap A_r)=$ Prob.of $r$ particular letters in right envelopes $=\frac{(n-r)!}{n!}$
$\therefore \sum p(A_1\cap A_2\cap .....\cap A_r)$
${=}^n{C}_r.\frac{(n-r)!}{n!}=\frac{1}{r!}$ where $r=1,2,3,....n.$
$\therefore \sum (\overline{A_1}\cap {\overline{A}}_2\cap \cdots \cap {\overline{A}}_n)$
$=1-\{\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\cdots +{(-1)}^n.\frac{1}{n!}\}$
which is equal to first $n-2$ terms in the expansion of $e^{-1}$.