Question

\[n\] letters to each of which correspond to an addressed envelope are placed in the envelope at random. What is the probability that no letter is placed in the right envelope?
A. \[1 - \left\{ {\dfrac{1}{{1!}} - \dfrac{1}{{2!}} + \dfrac{1}{{3!}} - ... + {{\left( { - 1} \right)}^n}.\dfrac{1}{{n!}}} \right\}\]
B. \[\left\{ {\dfrac{1}{{1!}} - \dfrac{1}{{2!}} + \dfrac{1}{{3!}} - ... + {{\left( { - 1} \right)}^n}.\dfrac{1}{{n!}}} \right\}\]
C. \[\left\{ {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ... + \dfrac{1}{{n!}}} \right\}\]
D. \[1 - \left\{ {\dfrac{1}{{1!}} + \dfrac{1}{{2!}} + \dfrac{1}{{3!}} + ... + \dfrac{1}{{n!}}} \right\}\]

Answer 1

Hint: In this question, we apply an axiomatic approach towards the probability distribution. We know, if the probability of an event occurring is say, p, then the probability that the event is not going to occur, i.e., the probability of its complement is (1-p). In this question too, we are going to use this fact of the complement of a probability. First, we are going to find the probability of the event that at least one letter is placed in the right envelope and then we are going to subtract it from 1 to find the required probability.

Complete step-by-step answer:
Let \[{A_i}\] denote the event that the ith letter is placed in the right envelope.
Then, our required probability is \[P\left( E \right) = \dfrac{{P\left( {{{\tilde A}_1} \cap {{\tilde A}_2} \cap ... \cap {{\tilde A}_n}} \right)}}{{P\left( {{A_1} \cup {A_2} \cup ... \cup {A_n}} \right)}}\]
Now, \[P\left( E \right) = 1 - P\left( {{A_1} \cup {A_2} \cup ... \cup {A_n}} \right)\]
 \[P\left( E \right) = 1 - \left[ {\sum {P\left( {{A_i}} \right) - \sum {P\left( {{A_i} \cap {A_j}} \right)} } } \right] + \sum {P\left( {{A_I} \cap {A_J} \cap {A_k}} \right)} - ... + {\left( { - 1} \right)^{n - 1}}P\left( {{A_1} \cap {A_2} \cap ... \cap {A_k}} \right)\]
Now, after the ith letter is placed in the correct envelope, the remaining letters can be placed in \[\left( {n - 1} \right)!\] ways. Hence,
 \[P\left( {{A_i}} \right) = \dfrac{{\left( {n - 1} \right)!}}{n}\]
Now, probability that r particular letters are in right envelopes, is:
 \[P\left( {{A_1} \cap {A_2} \cap ... \cap {A_r}} \right) = \dfrac{{\left( {n - r} \right)!}}{{n!}}\]
 \[\therefore \sum {P\left( {{A_1} \cap {A_2} \cap ... \cap {A_r}} \right){ = ^n}{C_r}.\dfrac{{\left( {n - r} \right)!}}{{n!}}} = \dfrac{1}{{r!}}\]
Hence, \[P\left( {{{\tilde A}_1} \cap {{\tilde A}_2} \cap ... \cap {{\tilde A}_n}} \right) = 1 - \left\{ {\dfrac{1}{{1!}} - \dfrac{1}{{2!}} + \dfrac{1}{{3!}} - ... + {{\left( { - 1} \right)}^n}.\dfrac{1}{{n!}}} \right\}\]
So, the correct answer is “Option A”.

Note: So, in questions like these, we first find the probability of the minimum of the possible event to occur. Then, we subtract that minimum event probability from 1, and we are going to have our answer. It is very important to remember two facts about probability – probability can never be less than 0, and probability can never be greater than 1. So, if we are having an answer which violates any of the two rules, then we need to recheck our method as something is wrong with the method or procedure being followed.