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There are four Envelopes with letters. Two are chosen Randomly and opened and found that they are wrongly addressed. Find the Probability that there are exactly two wrongly addressed envelopes.

My Try: Let the Envelopes be $E_1$,$E_2$,$E_3$ and $E_4$ and Corresponding Letters be $L_1$,$L_2$,$L_3$ and $L_4$ Since two opened are found wrongly addressed,implies there are minimum of two wrongly addressed envelopes.So Favorable cases are :

$1$.There are Exactly two wrongly addressed Envelopes i.e.,Remaining two are correctly addressed and this can happen in $\binom{4}{2}=6$ ways

$2$.There are exactly three wrongly addressed envelopes i.e., remaining one has correctly addressed and this an happen in $\binom{4}{3}\left(3!-1\right)=20 $ways

$3.$There are exactly four wrongly addressed envelopes and this an happen in $\left(4!-1\right)=23 $ways, so Required Probability is

$\frac{6}{6+20+23}=\frac{6}{49}$. I am not sure whether i have done in a right way, please help me if i am wrong.

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  • $\begingroup$ You have not mentioned any relation between the envelopes (i.e., if A received the wrong letter, then it doesn't mean that B or C or D received A's letter). If there is such relation, then you should explicitly mention it. $\endgroup$ – barak manos Apr 28 '14 at 11:24
  • $\begingroup$ This question is highly underspecified. For example, it can't be answered without a prior distribution on the assignments of letters to envelopes. The answer if the envelope assignments are drawn uniformly at random from all possibilities is different from the answer if the mail-addressing is expected to be highly reliable, so one failure is far more likely than two. $\endgroup$ – user2357112 Apr 28 '14 at 11:28
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If two randomly chosen envelopes are wrongly addressed, then the other two envelopes can only be correctly addressed if the two persons who receive the randomly chosen envelopes can get the letter that is meant for them by switching. Person $A$ must have received the letter for person $B$ and vice versa. If not then person $C$ or $D$ will receive a letter meant for $A$ or $B$ and more than $2$ letters are wrongly addressed.

Let's say that order $\left(1,2,3,4\right)$ stands for a fully correct addressing. The first two letters are wrongly addressed if we are dealing with for instance $\left(2,4,.,.\right)$. Counting tells you that there are $14$ of such cases. The first two letters are 'switched' in the cases $\left(2,1,3,4\right)$ and $\left(2,1,4,3\right)$ and in only case $\left(2,1,3,4\right)$ there are exactly $2$ wrongly addressed letters. If the first two letters are not switched then there are more than $2$ wrongly addressed letters as argued above. We find a probability of $\frac{1}{14}$ that there are exactly $2$ letters are wrongly addressed under condition that the first two are wrongly addressed.

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  • $\begingroup$ Nothing really says that there were four addresses on the letters and the same four addresses on the envelopes, and that the letters have been mixed up. They could be letters ABCD sent to addresses XYZV. $\endgroup$ – gnasher729 Apr 28 '14 at 10:24
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    $\begingroup$ @gnasher729 It is indeed not mentioned in the question, but I presumed it. If this presumption is not correct then most probably the question would not have been asked. $\endgroup$ – drhab Apr 28 '14 at 10:32
  • $\begingroup$ @gnasher729, yes even i presumed that there are 4 addresses on letters and same four on envelopes. $\endgroup$ – Ekaveera Kumar Sharma Apr 28 '14 at 11:39
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We want to find the probability of exactly two wrongly addressed envelopes in 4 envelopes, when we already knew that 2 of them are wrongly addressed (known as conditional probability). So we can omit "chosen randomly" from your problem.

From this point of view, we need to find the probability of being correctly addressed for the remaining two envelopes.

Without loss of generality consider E1 and E2 as the ones which have been opened (not by us). We have two cases:

  • If L2 is in E1, there is 3 ways to fill E2, afterward there is 2 ways to fill E3 and finally 1 way to fill E4. So (by using Rule of Product) there is $1\times 3\times 2\times 1=6$ ways for E1 and E2 to be wrongly addressed in this case.
  • If L2 is not in E1, then there is 2 ways to fill E1, afterward there is 2 ways to fill E2, afterward there is 2 ways to fill E3 and finally 1 way to fill E4. So there is $2\times 2\times 2\times 1 = 8$ ways for E1 and E2 to be wrongly addressed in this case.

In these $6+8$ ways, there is only one way in which E3 and E4 are correctly addressed.

So the answer would be $\frac{1}{6+8}$.

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The problem is not quite clear. First, do we have four letters, four envelopes with correct addresses, and mixed up the letters putting them into the envelopes (in that case having exactly one envelope with the wrong address would be impossible), or do we have letters in envelopes, and sometimes the wrong address was written on the envelope? Second, are the errors independent? Let's say I put four letters in a row, four envelopes in a row, then the first envelope drops on the floor and I put it back up at the end of the row instead of at the beginning of the row. Then all envelopes would be wrongly addressed.

For simplicity, let's assume that with independent probability p I write an incorrect address on an envelope. We can calculate the probability that two random envelopes have the wrong address, that exactly two out of all four envelopes have the wrong address, and the conditional probability that exactly two have the wrong address if two random ones had the wrong address. Actually, the probability for two random ones wrong is p^2, the probability for all others correct is (1 - p)^2. However, we don't know p.

If we had the choice of picking a careful or careless person to address the letters, and know how their "carefulness" is distributed, then we can from the fact that we did have two letters wrongly addressed conclude that the person we picked is a bit on the careless side, and the probability that the others are both correct is likely lower. Say we picked randomly between two people, one with probability 0.1 of addressing a letter wrong, one with probability 0.9 of getting it wrong - having two random letters addressed wrongly makes it much more likely that the second person addressed them, and the other two are wrong as well.

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Assumption: The letters all have different addresses, and these four different addresses were pre-printed onto the envelops correctly, so that the mismatch occurs during parcelling the letters into the envelops. Also there is one letter placed in each envelope.

Let $A,B,C,D$ be the envelopes' addresses, with $a,b,c,d$ being the letters' addresses.

We have opened 2 envelopes, let's call them $C,D$, and given that these contain the wrong letters we wish to calculate the conditional probability that the remaining two envelopes, $A,B$, correctly address their contents.

So we want to find: $p=\mathrm{P}(A=a, B=b \mid C\neq c, D\neq d)$

By definition of conditional probability: $$p = \frac{\mathrm{P}(A\!=\!a, B\!=\!b, C\!\neq\!c, D\!\neq\!d)}{\mathrm{P}(C\!\neq\!c, D\!\neq\!d)}$$ By mutual exclusivity of the marginal events, $\mathrm{P}((X,Z)\cup( Y,Z))=\mathrm{P}(X,Z)+\mathrm{P}(Y,Z)$: $$p = \frac{\mathrm{P}(A\!=\!a, B\!=\!b, C\!\neq\!c, D\!\neq\!d)}{\mathrm{P}((C\!=\!a \cup C\!=\!b), D\!\neq\!d)+\mathrm{P}(C\!=\!d \cap D\!\neq\!d)}$$ By expanding using conditional probability: $$p = \frac{\mathrm{P}(A\!=\!a)\cdot\mathrm{P}(B\!=\!b\mid A\!=\!a)\cdot\mathrm{P}(C\!\neq\!c, D\!\neq\!d\mid A\!=\!a,B\!=\!b)}{\mathrm{P}(C\!=\!a \cup C\!=\!b)\cdot\mathrm{P}(D\!\neq\!d\mid C\!=\!a \cup C\!=\!b)+\mathrm{P}(C\!=\!d)\cdot\mathrm{P}(D\!\neq\!d\mid C\!=\!d)}$$ Assigning values by assuming the letters were somehow assigned to envelops by unbiased sorting. $$p = \frac{\frac 14\cdot\frac 13\cdot\frac 12}{\frac 24\cdot\frac 23 + \frac 14\cdot\frac 11}$$ Evaluating: $$p = \frac{1}{14}$$

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The key to this is the number of derangements:

  • Two letters can go in wrong envelopes $1$ way.
  • Three letters can go in wrong envelopes $2$ ways.
  • Four envelopes can go in wrong envelopes $9$ ways.

Given that you have looked in two envelopes and found they are wrong, you then want to consider whether the others may be correct, giving $1 \times {2 \choose 2}=1$, $2 \times {2 \choose 1}=4$, and $9 \times {2 \choose 0}=9$.

Your three numbers should be $1$, $4$ and $9$ respectively (or if you want to reflect "two are chosen randomly" in full then multiply by six to give $6$, $24$ and $54$ respectively, but this is just a scaling factor). So the probability is $\frac{1}{1+4+9}=\frac{1}{14}$.

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In a pack of four envelopes,

  1. one pair of envelopes can be wrongly addressed in only 1 way. But these two envelopes can be chosen from four envelopes in $^{4}C_{2}$ ways. Hence exactly two envelopes can be wrongly addressed in $^{4}C_{2} * 1 = 6$ ways.
  2. one triplet of envelopes can be wrongly addressed in 2 ways. Suppose 1-2-3-4 is the correctly addressed sequence and supposing that envelope number 4 is always correctly addressed, the 2 ways of wrongly addressing the remaining envelopes are 3-1-2-4 and 2-3-1-4. But any randomly chosen three envelopes can be wrongly addressed. Hence exactly three envelopes can be wrongly addressed in $^{4}C_{3} * 2 = 8$ ways.
  3. similarly, one quartet of envelopes can be wrongly addressed in 9 ways. Suppose 1-2-3-4 is the correctly addressed sequence, the envelopes can be wrongly addressed as 2-1-4-3, 2-3-4-1, 2-4-1-3 and so on. But four envelopes can be randomly chosen in $^{4}C_{4}$ ways. Hence exactly four envelopes can be wrongly addressed in $^{4}C_{4} * 9 = 9$ ways.

Therefore, the probability that there are exactly two wrongly addressed envelopes = $\frac{6}{6 + 8 + 9}$ i.e. 26.09$\%$.

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  • $\begingroup$ You need to assess this conditional on the knowledge that two envelopes have the wrong address. $\endgroup$ – Graham Kemp Apr 29 '14 at 1:46

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