[Math] Probability that there are exactly two wrongly addressed envelopes

combinatoricsprobability

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.

Best Answer

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.