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A secretary writes four letters and the corresponding addresses on envelopes. If he inserts the letters in the envelopes at random irrespective of the addresses, (i) find the probability that only one letter is placed in the corresponding envelope, (ii) calculate the probability that all the letters are are wrongly placed. (iii) find the probability that only two letters are placed in the corresponding envelopes.

Please someone help me to solve the problem.

My attempt

Let $A_i$ ($i=1,2,3,4$) be the event that $i$th letter is placed to $i$th envelop.

Now the probability that atleast one letter is placed to the correct envelop = $P(A_1+A_2+A_3+A_4)=\sum A_i -\sum{A_iA_j}+\sum{A_iA_jA_k} -A_1A_2A_3A_4$. Now, $P(A_i)=\frac{1}{4}, P(A_iA_j)=P(A_i)P(A_j/A_i)=\frac{1.1}{4.3}=\frac{1}{12}$. Similarly, $P(A_iA_jA_k)=\dots=1/24$ and $P(A_1A_2A_3A_4)=\dots=1/24$. Therefore $P(A_1+A_2+A_3+A_4)=\sum A_i -\sum{A_iA_j}+\sum{A_iA_jA_k} -A_1A_2A_3A_4$ $=4.\frac{1}{4}-6.\frac{1}{12}+4.\frac{1}{24}-\frac{1}{24}=\frac{5}{8}$.

  • Part (ii)

Prob that all letters are wrongly placed = $P(\bar{A_1}\bar{A_2}\bar{A_3}\bar{A_4})=1-P(A_1+A_2+A_3+A_4)=1-\frac{5}{8}=\frac{3}{8}$.

Please help me to solve the part (i) & (iii). Please solve the problem using chain rule of probability as I have solved for the second part.

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  • $\begingroup$ Since $A_1,\ldots,A_4$ are events, it does not make to sum it. Use intersection or union of events instead. $\endgroup$
    – iiivooo
    Commented Mar 5, 2015 at 11:03
  • $\begingroup$ @iiivooo + stands for union and product stands for intersection. Its a general convention. $\endgroup$ Commented Mar 5, 2015 at 11:05
  • $\begingroup$ Please someone help me to solve the problem. $\endgroup$ Commented Mar 6, 2015 at 6:50

1 Answer 1

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$$\newcommand{\c}[2]{^{#1}{\rm C}_{#2}}\newcommand{\d}[1]{{\rm #1}} a)\frac{\c41.!3}{4!}=\frac13\\b)\frac{!4}{4!}=\frac38\\c)\frac{\c42.!2}{4!}=\frac14$$


Explanation

  • In a) Choose 1 out of four to be correctly placed and derange rest (make them go into wrong places). Total $4!$

  • In b) Derange all (All wrong places). Total $4!$

  • In c) Choose 2 out of four to be correctly placed and derange rest (make them go into wrong places). Total $4!$


Alternate(without $!n$)

  • a) "the probability that only one letter is placed in the corresponding envelope":

$$\rm P\left(\sum A_i\overline{A_jA_kA_l}\right)=4P(A_1\overline{A_2A_3A_4})$$

Now, $$\rm P(A_1\overline{A_2A_3A_4})=P(A_1)P(\overline{A_2A_3A_4}\mid A_1)=\frac14P(\overline{A_2A_3A_4}\mid A_1)$$

Now, $$\rm P(\overline{A_2A_3A_4}\mid A_1)\\=1-P(A_2+A_3+A_4)=1-\left[\sum P(A_i)-\sum P(A_iA_j)+\sum P(A_iA_jA_k)\right]\\=1-\left[3.\frac13-3.\frac13.\frac12+\frac13.\frac12.\frac11\right]=1/3$$ Hence, $$\rm P\left(\sum A_i\overline{A_jA_kA_l}\right)=1/3$$

  • b) You already did that.($5/8$), anyways it's done both ways now.

  • c) "the probability that only two letters are placed in the corresponding envelopes":

Similarly: $$\rm P\left(\sum A_iA_j\overline{A_kA_l}\right)=\;\c42P(A_1A_2\overline{A_3A_4})=6P(A_1A_2)P(\overline{A_3A_4}\mid A_1A_2)\\=6.\left[\frac14.\frac13\right].\left[1-\left(2.\frac12-\frac12.\frac11\right)\right]=\frac14$$

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  • $\begingroup$ math.stackexchange.com/users/67609/adg Please explain in full in support of a) and c). What is the meaning of the notation !n? $\endgroup$ Commented Mar 6, 2015 at 8:36
  • $\begingroup$ Please comment at your earliest. $\endgroup$ Commented Mar 6, 2015 at 11:40
  • $\begingroup$ @user1942348 it's called derangement(!n),i.e. rearranging n objects (already placed on a line) in such a manner that they don't go to their respective places.BTW edited $\endgroup$
    – RE60K
    Commented Mar 6, 2015 at 17:43
  • $\begingroup$ How to then one can evaluate !3, !4, ... !n ? $\endgroup$ Commented Mar 6, 2015 at 18:41
  • $\begingroup$ @user1942348 $!n=n!*[1/0!-1/1!+1/2!-1/3!+1/4!-1/5!.....+(-1)^n/n!]$ $\endgroup$
    – RE60K
    Commented Mar 6, 2015 at 18:47

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