Need to calculate the probability In an office, after having a very busy day, the secretary is just leaving the seat when the boss calls her and hands over the drafts for four letters with addresses. The secretary types the letters and addresses on envelopes. But when she is going to put letters in envelopes, she puts letters in envelopes randomly without seeing the corresponding addresses. What is the probability that exactly three letters will be dispatched to correct addresses?
I'm new to this so I'm a bit confused.
As far as I know P(event) = outcomes that meet our criteria / all possible outcomes
So, what I did is:
P(3) = 3 / 4 but I'm not fully convinced if this is correct. I'm just not sure.
 A: The probability is $0,$ since it is impossible for her to put exactly three of the four letters in the right envelope. If three of the letters are in the right envelope, then there's only one envelope left for the fourth letter.
$$
\begin{array}{l}
A\left\{\begin{array}{l}
AB\left\{
\begin{array}{l}
ABCD & & 4 \text{ matches} \\
ABDC & & 2 \text{ matches}
\end{array}
\right. \\[8pt]
AC
\left\{
\begin{array}{l}
ACBD & & 2 \text{ matches} \\
ACDB & & 1 \text{ match}
\end{array}
\right. \\[8pt]
AD
\left\{
\begin{array}{l}
ADBC & & 1 \text{ match} \\
ADCB & & 2 \text{ matches}
\end{array}
\right.
\end{array}
\right. \\[10pt]
B \left\{
\begin{array}{l}
BA\left\{
\begin{array}{l}
BACD & & 2 \text{ matches} \\
BADC & & 0 \text{ matches}
\end{array}
\right. \\[8pt]
BC
\left\{
\begin{array}{l}
BCAD & & 1 \text{ match} \\
BCDA & & 0 \text{ matches}
\end{array}
\right. \\[8pt]
BD
\left\{
\begin{array}{l}
BDAC & & 0 \text{ matches} \\
BDCA & & 1 \text{ match}
\end{array}
\right.
\end{array}
\right. \\[10pt]
C\left\{\begin{array}{l}
CA\left\{
\begin{array}{l}
CABD & & 1 \text{ match} \\
CADB & & 0 \text{ matches}
\end{array}
\right. \\[8pt]
CB
\left\{
\begin{array}{l}
CBAD & & 2 \text{ matches} \\
CBDA & & 1 \text{ match}
\end{array}
\right. \\[8pt]
CD
\left\{
\begin{array}{l}
CDAB & & 0 \text{ matches} \\
CDBA & & 0 \text{ matches}
\end{array}
\right.
\end{array}
\right. \\[10pt]
D \left\{
\begin{array}{l}
DA\left\{
\begin{array}{l}
DABC & & 0 \text{ matches} \\
DACB & & 1 \text{ match}
\end{array}
\right. \\[8pt]
DB
\left\{
\begin{array}{l}
DBAC & & 1 \text{ match} \\
DBCA & & 2 \text{ matches}
\end{array}
\right. \\[8pt]
DC
\left\{
\begin{array}{l}
DCAB & & 0 \text{ matches} \\
DCBA & & 0 \text{ match}
\end{array}
\right.
\end{array}
\right.
\end{array}
$$
A: Remember that in this case you are not measuring one probability. Each time the secretary puts a letter into an envelope that is a separate action that has its own probability so usually in order to measure the probability of multiple actions you multiply all of the separate probabilities of the separate actions together. Moreover this example is a probability without replacement which means the probability of each action depends on the previous actions. So, the probability of the first time putting the correct letter in the correct envelope is 1/4 because there are 4 envelopes and only one correct 1. The probability of the second time they match is 1/3 because you already put one letter in an envelope meaning there is 3 envelopes left of which one is the correct one. So if wanted to predict the probability of 2 envelopes matching the correct address we would multiply (1/4)*(1/3) to get a probability of 1/12. However, if we continue this then when we get to the last envelope the probability of that envelope and letter matching is 1/1 meaning that if three envelopes are chosen correctly the final envelope also has to be correct since individual probability is 100 percent. This means it is impossible for the first three envelopes to be correct without the final one also being correct.
