Probability that there are exactly two wrongly addressed envelopes 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.
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.
In a pack of four envelopes, 


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.
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.
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$\%$.