Probability ( Letters and Envelopes ).

A secretary types three letters and the three corresponding envelopes. In a hurry , he places at random one letter in each envelope. We need to find the probability that at least one letter is in the correct envelope.

The way I figured it out : The given probability can be written as :

1 - ( Probability that no letter is in the correct envelope )

that is : $1\: -\: \frac{2}{3}* \frac{1}{2}* 1$

Explanation for this : For the first letter only one envelope is correct , so probability of getting a wrong envelope becomes $\frac{2}{3}$ , for the second letter , there are 2 options , one is correct other is incorrect so Probability of getting the wrong one is $\frac{1}{2}$ , and for the last one there is only one envelope which is obviously the wrong one so $1$.

Is the above solution correct ??

Let us have n letters corresponding to which there exist n envelopes bearing different addresses.

A Match occurs if $Letter_{i}$ gets into $Envelope_{i}$ , $Letter_{j}$ gets into $Envelope_{j}$. Let us denote this event as $E_{i}$ .

A NO Match occurs if $Letter_{i}$ gets into $Envelope_{j}$ , $Letter_{j}$ gets into $Envelope_{i}$. Let us denote this event as $\overline{E_{i}}$ .

$E_{i}$ : Denote the Event where that the ith object occupies the ith position corresponding to its number. Then, the probability 'p' that P(None of the objects occupies the place corresponding to its number) = P( No Letter will be in its proper Envelope corresponding to its number) is given by : $ p = P(\overline{E1} \cap \overline{E2} \cap \overline{E3}.... \cap \overline{En} ) = 1 - P(\text{Atleast one of the objects occupies the place corresponding to its number})= 1 - P(\text{Atleast One Letter will be in its proper Envelope corresponding to its number})= 1 - P(E1 \cup E2\cup E3.... \cup En) = 1 - [\sum_{i=1 }^{n}P(E_{i}) - \sum_{i=1 }^{n}\sum_{j=1 }^{n}P(E_{i} \cap E_{j})....+(-1)^{n-1}P(E_{1} \cap E_{2}\cap E_{3}....... \cap E_{n}) ]= 1 - [\frac{\binom{n}{1}}{n} - \frac{\binom{n}{2}}{n(n-1)} + \frac{\binom{n}{3}}{n(n-1)(n-2)} - ..... + \frac{(-1)^{n-1}}{n(n-1)(n-2)...3.2.1} ] = 1- [ 1- \frac{1}{2!} + \frac{1}{3!}-...+\frac{{(-1)^{(n-1)}}}{n!}]= \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - ........ + \frac{(-1)^{(n)}}{n!} = \sum_{k = 0 }^{n}\frac{(-1)^{k}}{k!}$ ......................................................................

But for large n :

p = $1-1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} - ........... = e^{-1}$

And, the Probability of Atleast One match : $1 - p = (1-e^{-1})$

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Therefore, P(None of the n letters goes to the correct envelope ) = $\sum_{k = 0 }^{n}\frac{(-1)^{k}}{k!}$

Therefore, $P(\text{Atleast One Letter will be in its proper Envelope corresponding to its number})= [ 1- \frac{1}{2!} + \frac{1}{3!}-...+\frac{{(-1)^{n-1}}}{n!}]$