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There are four envelopes and corresponding $4$ letters.If the letters are placed in the envelopes at random,what is the probability that all the letters are not placed in the right envelopes?


Approach-

P(all letters in place) = 1/4!

P(required) = 1-(P letters not in place) = 1-1/4!

But the problem is that P(required) does include all the arrangements in which all are out of place but also the arrangements in which just 2 are out of place.How to I find the real P(required) in which only the all four are displaced

NOTE = the textbook gives answer 1-1/4! but I am not sure if its correct.

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    $\begingroup$ The probability that there is at least one letter which is not in the correct envelope is indeed $1-\frac{1}{4!}$. A more interesting problem is the probability that all the letters are in a wrong envelope. This is the problem of derangements (see Wikipedia). $\endgroup$ – André Nicolas Feb 3 '14 at 20:01
  • $\begingroup$ indeed the question says that the probability to find out is ALL not in place,so i can safely assume that the ans in textbook is wrong $\endgroup$ – user2688772 Feb 3 '14 at 20:05
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    $\begingroup$ It is a matter of exact wording: "not all in place" does not mean the same thing as "all not in place." One would need to know the precise question, in the original language if it is not English. $\endgroup$ – André Nicolas Feb 3 '14 at 20:08
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    $\begingroup$ Let us interpret the question as asking for the probability that every letter is in an incorrect envelope. (The wording of the question is too imprecise to force that interpretation.) Then we can use the Wikipedia reference I made earlier. Or we can compute directly. Letter $1$ can be in envelope $2$, $3$, or $4$ ($3$ choices). For the choice envelope $2$, calculate by explicit listing the number of ways to finish. There will be $3$, so the total number of arrangements where all are wrong is $(3)(3)$, the probability is $\frac{9}{4!}$. $\endgroup$ – André Nicolas Feb 3 '14 at 20:39
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    $\begingroup$ What is the title and authors of your textbook? $\endgroup$ – user940 Feb 3 '14 at 21:14
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The intended question is not clear from the wording. What does "what is the probability that all the letters are not placed in the right envelopes" mean? If it said "that not all the letters are placed in the right envelope," then the question would be reasonably clear, the answer is $1-\frac{1}{4!}$. If it said "that every letter is placed in a wrong envelope," then we would be looking at the problem of counting derangements.

In the cited article, it is shown that the number of derangements of $n$ objects is $n! \sum_{i=2}^{n}\frac{(-1)^i}{i!}$. In our $n=4$ case, the probability is therefore $\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}$, which is $\frac{3}{8}$.

For $n=4$, we can also count the derangements by more or less explicit listing. Let the letters be $L_1$ to $L_4$, and the envelopes be $E_1$ to $E_4$.

Letter $L_1$ can be placed in any of $E_2, E_3,E_4$. Each placement gives the same number of derangements, so we find the number of derangements in which $L_1$ is placed in $E_2$, and multiply by $3$.

If $L_1$ is placed in $E_2$, then either (i) $L_2$ is placed in $E_1$ or (ii) $L_2$ is placed in $E_3$ or $E_4$. It is easy to see that in Case (i), $L_3$ must be in $E_4$, and $L_4$ in $E_1$. In the same way, each of the choices in Case (ii) gives a unique derangement. So there are $3$ derangements that take $L_1$ to $E_2$, and therefore a total of $(3)(3)$ derangements. It follows that the probability that every letter is in a wrong envelope is $\frac{9}{4!}$.

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