Determine the required value of the missing probability to make the distribution a discrete probability distribution. \begin{tabular}{cc} \hline $\mathbf{x}$ & $\mathbf{P}(\mathbf{x}) \square$ \\ \hline 3 & 0.32 \\ \hline 4 & $?$ \\ \hline 5 & 0.27 \\ \hline 6 & 0.12 \\ \hline \end{tabular} \[ P(4)=\square \text { (Type an integer or a decimal.) } \]

Step 1 ：Given the probabilities for x = 3, 5, and 6 as 0.32, 0.27, and 0.12 respectively.

Step 2 ：The sum of all probabilities in a probability distribution must equal 1.

Step 3 ：Therefore, to find the missing probability for x = 4, we need to subtract the sum of the given probabilities from 1.

Step 4 ：Calculate the sum of the given probabilities: \(0.32 + 0.27 + 0.12 = 0.71\)

Step 5 ：Subtract this sum from 1 to find the missing probability: \(1 - 0.71 = 0.29\)

Step 6 ：Final Answer: The required value of the missing probability to make the distribution a discrete probability distribution is \(\boxed{0.29}\).