Full house hand

The probability is:

$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{2}}{\binom{52}{5}}=\frac{3744}{2598960}\approx \boxed{0.001441}$

The denominator is a count of the number of ways to select $5$ cards from a set of $52$ : $\binom{52}{5}$

The numerator can be interpreted as follows:

First, choose $1$ rank from among the $13$ : $\binom{13}{1}$

Then, choose $3$ cards from among the $4$ cards of that rank: $\binom{4}{3}$

Next, choose another $1$ rank from the remaining $12$ ranks: $\binom{12}{1}$

Finally, choose $2$ cards from among the $4$ cards of that rank $\binom{4}{2}$

Then, use the rule of product to find the total number of distinct hands which are full houses.

$P(\text{Full House}) = \dfrac{\textcolor{#3D99F6}{^{13}P_2} . \textcolor{#CEBB00}{^4C_3} . \textcolor{#D61F06}{^4C_2}}{\textcolor{#20A900}{^{52}C_5}} = \boxed{0.001441}$

$\textcolor{#3D99F6}{\text{Selecting any 2 ranks}}$

$\textcolor{#CEBB00}{\text{Choosing 3 out of 4 cards}}$

$\textcolor{#D61F06}{\text{Choosing 2 out of 4 cards}}$

$\textcolor{#20A900}{\text{Choosing 5 out of 52 cards}}$