91Ó°ÊÓ

Define four events, for $i=1,2,3,4$ :

$E_i=$ event that $i$-th color is absent from the $13$ cards.

$E_i=$event that $i$-th color is absent from the 13 cards

These events have equal probabilities, and the wanted probability is

$P\left[\left(E_1∪E_2∪E_3∪E_4\right)\right]$

Remember the Proposition 4.4. (the law of inclusion and exclusion):

Let $A_1,A_2,…,A_n$be some events.

localid="1650016733092" $$\begin{gathered} P\left(A_1∪A_2,…A_n\right)=\underset{i∈\{1,2,…,n\}}{∑}P\left(A_i\right)+(-1{)}^1\underset{i_1<i_2}{∑}P\left(A_{i_1}âˆ\copyright A_{i_2}\right) \\ +…+(-1{)}^{n-1}\underset{i_1<i_2<…<i_n}{∑}P\left(A_{i_1}âˆ\copyright A_{i_2}…âˆ\copyright A_{i_n}\right) \\ {i}_1,i_2,…i_n∈\{1,2,…,n\} \end{gathered}$$

The probabilities that occur in the inclusion and exclusion equations are now calculated. The probability of any two or three events intersecting is the same since the events are symmetrical. There are no hands that are devoid of all four hues.

$$\begin{gathered} P\left(E_i\right)=\frac{\left(\begin{array}{l}39\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)} \\ P\left(E_{i_1}âˆ\copyright E_{i_2}\right)=\frac{\left(\begin{array}{c}26\\ 13\end{array}\right)}{\left(\begin{array}{c}52\\ 13\end{array}\right)} \\ P\left(E_{i_1}âˆ\copyright E_{i_2}âˆ\copyright E_{i_3}\right)=\frac{\left(\begin{array}{l}13\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)} \\ P\left(E_1âˆ\copyright E_2âˆ\copyright E_3âˆ\copyright E_4\right)=0 \\ {i}_1,i_2,i_3∈\{1,2,3,4\} \end{gathered}$$

And $k$indexes out of $n$(which is the same as $k$ordered indexes) can be any of the $\left(\begin{array}{l}n\\ k\end{array}\right)$combinations.

localid="1650016750162" $$\begin{gathered} P\left(E_1∪E_2∪E_3∪E_4\right)=\left(\begin{array}{l}4\\ 1\end{array}\right)P\left(E_1\right)+(-1{)}^1\left(\begin{array}{l}4\\ 2\end{array}\right)P\left(E_1âˆ\copyright E_2\right)+(-1{)}^2\left(\begin{array}{l}4\\ 3\end{array}\right)P\left(E_1âˆ\copyright E_2âˆ\copyright E_3\right)+(-1{)}^3\left(\begin{array}{l}4\\ 4\end{array}\right)P\left(E_1âˆ\copyright E_2âˆ\copyright E_3âˆ\copyright E_4\right) \\ =4P\left(E_1\right)-6P\left(E_1âˆ\copyright E_2\right)+4P\left(E_1âˆ\copyright E_2âˆ\copyright E_3\right)-P\left(E_1âˆ\copyright E_2âˆ\copyright E_3âˆ\copyright E_4\right) \\ =4\frac{\left(\begin{array}{l}39\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}-6\frac{\left(\begin{array}{l}26\\ 13\end{array}\right)}{\left(\begin{array}{l}52\\ 13\end{array}\right)}+4\frac{1}{\left(\begin{array}{l}52\\ 13\end{array}\right)}-0 \\ ≈0.051 \end{gathered}$$