<s>[INST] probabilty problem how to solve Six cards are drawn with replacement form on ordinary deck. What is the probabilty that each of four suits will be represented at least once among the six cards?
 [/INST] Consider the different cases:
1 suit
$$
\begin{align*}
P(\text{Only } \clubsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
P(\text{Only } \heartsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
P(\text{Only } \spadesuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
P(\text{Only } \diamondsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6
\end{align*}
$$
So 4 cases.
2 suits
$$
\begin{align*}
P(\text{Only } \clubsuit \text  { AND } \heartsuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 2\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
&.\\&.\\&.\\
\end{align*}
$$
The idea here is you EXCLUDE the cases where only $\clubsuit$ or $\heartsuit$ are chosen. The residual probability is the probability of a combination of both $\clubsuit$ and $\heartsuit$
You will realize that there are ${4 \choose 2} = 6$ cases ie
$$
(\clubsuit,\heartsuit) (\clubsuit,\spadesuit) (\clubsuit,\diamondsuit) \\
(\heartsuit,\spadesuit) (\heartsuit,\diamondsuit) (\spadesuit,\diamondsuit) 
$$
So do your multiplication accordingly.
3 suits
$$
\begin{align*}
P(\text{Only } \clubsuit \text  { AND } \heartsuit { AND } \spadesuit \text{ suit is drawn}) &= \begin{pmatrix}\frac{3}{4}\end{pmatrix}^6 - 3\times\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 3\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\\
&.\\&.\\&.\\
\end{align*}
$$
Similarly to 2 suits, the idea is to exclude the case where only 2 suits are selected, and only 1 suit is selected. There are $3$ ways that only 2 suits are selected and $3$ ways that only 1 suit can be selected.
There are 4 cases, namely
$$
(\clubsuit,\heartsuit, \spadesuit) (\heartsuit, \spadesuit, \diamondsuit) \\
(\spadesuit,\diamondsuit, \clubsuit) (\diamondsuit,\clubsuit, \heartsuit)
$$
So do the addition, and take compliment.
$$
\begin{align*}
\text{Req. Probability} &=
1\\&-
4\begin{bmatrix}\begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}\\&-
6\begin{bmatrix}\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 - 2\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}\\&-
4\begin{bmatrix}\begin{pmatrix}\frac{3}{4}\end{pmatrix}^6 - 3\times\begin{pmatrix}\frac{2}{4}\end{pmatrix}^6 + 3\times \begin{pmatrix}\frac{1}{4}\end{pmatrix}^6\end{bmatrix}
\end{align*}
$$
</s>