Question

If n letters are placed into n addressed envelopes at random, the probability that at least one letter will go into a wrongly addressed envelope is
\[\begin{align}
  & A.\dfrac{1}{n} \\
 & B.\dfrac{n-1}{n} \\
 & C.1-\dfrac{1}{n!} \\
 & D.\dfrac{1}{n!} \\
\end{align}\]

Answer 1

Hint: At first, count the total number of outcomes which is total permutations of placing envelopes which is n! Then, finding the number of ways of placing correct letters to envelopes and then complement them from the set. Then, use the formula of probability which is, $P\left( E \right)=\dfrac{n\left( F \right)}{n\left( E \right)}$ where P (E) is probability, n (F) total number of favorable outcomes and n (E) total number of outcomes for an event.

Complete step by step answer:
In the question, we are said that ‘n’ letters are placed into n addressed envelope at random and we have to find the probability for which one letter will go into a wrongly addressed envelope.
Before finding let us know what probability is.
Probability is the branch of mathematics concerning the numerical description of how likely an event to occur or how likely it is that proposition is true. A simple example is the tossing of a fair or an unbiased coin. The probability of an event is number between 0 and 1, where roughly speaking, 0 indicates the impossibility of the event, and 1 indicates its certainty. The higher the probability of an event the more likely it is that event will occur. Since, the coin be fair the two outcomes "heads" and "tails" are both equally probable, the probability of heads equal to the probability of tails and since no other outcomes are possible, the probability of either heads or tails is $\dfrac{1}{2}$
Generally, we find probability using the formula,
\[P\left( E \right)=\dfrac{n\left( F \right)}{n\left( E \right)}\]
Here P (E) represents the probability of the event, n (F) represents a total number of favorable outcomes of the event and n (E) represents a total number of outcomes for an event.
Now, we know that, the total number of permutations or ways of placing letters are n!
Also, the number of ways in which all letters are placed in the correct envelope is 1.
So for at least 1 letter to be misplaced to compliment the set which can be done in n!-1 ways.
Hence, we know the total number of events or n (E) are n! and the number of favorable outcome or n(F) are (n!-1) so on substituting we get, probability or P(E) as,
\[\begin{align}
  & P\left( E \right)=\dfrac{n\left( F \right)}{n\left( E \right)} \\
 & \Rightarrow \dfrac{n!-1}{n!} \\
 & \Rightarrow 1-\dfrac{1}{n!} \\
\end{align}\]
Thus, the correct option is C.

Note:
Generally while calculating probability, students tend to miss out on any favorable outcome or make mistakes while calculating the total number of outcomes which can make their answer incorrect. So, they should be careful. For example, here we have probability as \[\dfrac{n!-1}{n!}\] , but students might take it as \[\dfrac{1}{n!}\] and then choose option d as the correct answer.