Determine the required value of the missing probability to make the distribution a discrete probability distribution.
\begin{tabular}{cc}
\hline $\mathbf{x}$ & $\mathrm{P}(\mathbf{x})$ \\
\hline 3 & 0.21 \\
\hline 4 & $?$ \\
\hline 5 & 0.23 \\
\hline 6 & 0.29 \\
\hline
\end{tabular}
$P(4)=$
(Type an integer or a decimal.)

Step 1 ：The problem is asking to find the missing probability value for the discrete probability distribution. In a probability distribution, the sum of all probabilities must equal 1.

Step 2 ：We are given the probabilities for the values 3, 5, and 6. We can denote these as p3 = 0.21, p5 = 0.23, and p6 = 0.29.

Step 3 ：To find the missing probability, denoted as p4, we subtract the sum of the given probabilities from 1. This can be expressed as \(p4 = 1 - (p3 + p5 + p6)\).

Step 4 ：Substituting the given values into the equation, we get \(p4 = 1 - (0.21 + 0.23 + 0.29)\).

Step 5 ：Solving the equation, we find that \(p4 = 0.27\).

Step 6 ：So, the missing probability value for the discrete probability distribution is \(\boxed{0.27}\).