Updated the questions with fixes(added new supported notations)

Author: Jeet009

questions/185_pmf_normalization_constant 2/learn.md

@@ -1,11 +1,11 @@
 ## Learning: PMF normalization constant
 
 ### Idea and formula
-- **PMF requirement**: A probability mass function must satisfy ∑ p(xᵢ) = 1 and p(xᵢ) ≥ 0.
-- **Normalization by a constant**: If probabilities are given up to a constant, you determine that constant by enforcing the sum-to-1 constraint.
-  - If the form is p(xᵢ) = K · wᵢ with known nonnegative weights wᵢ, then
-    - ∑ p(xᵢ) = K · ∑ wᵢ = 1 ⇒ **K = 1 / ∑ wᵢ**.
-  - If the given expressions involve K in a more general way (e.g., both K and K² terms), still enforce ∑ p(xᵢ) = 1 and solve the resulting equation for K. Choose the solution that makes all probabilities nonnegative.
+- **PMF requirement**: A probability mass function must satisfy $\sum_i p(x_i) = 1$ and $p(x_i) \ge 0$.
+- **Normalization by a constant**: If probabilities are given up to a constant, determine that constant by enforcing the sum-to-1 constraint.
+  - If the form is $p(x_i) = K\,w_i$ with known nonnegative weights $w_i$, then
+    - $\sum_i p(x_i) = K \sum_i w_i = 1 \Rightarrow$ $\displaystyle K = \frac{1}{\sum_i w_i}$.
+  - If the given expressions involve $K$ in a more general way (e.g., both $K$ and $K^2$ terms), still enforce $\sum_i p(x_i) = 1$ and solve the resulting equation for $K$. Choose the solution that makes all probabilities nonnegative.
 
 ### Worked example (this question)
 Suppose the PMF entries are expressed in terms of K such that, when summed, the K-terms group as follows:
@@ -23,19 +23,28 @@ One concrete way to realize this is via the following table of outcomes and prob
 | x₄ | 3K + 7K²     |
 | x₅ | K            |
 
-These add up to 9K + 10K² as required.
+These add up to $9K + 10K^2$ as required.
 
 Enforce the PMF constraint:
 
-- 9K + 10K² = 1 ⇒ 10K² + 9K − 1 = 0
+$$
+9K + 10K^2 = 1 \;\Rightarrow\; 10K^2 + 9K - 1 = 0
+$$
 
 Quadratic formula reminder:
 
-- For aK² + bK + c = 0, the solutions are K = [−b ± √(b² − 4ac)] / (2a).
+$$
+\text{For } aK^2 + bK + c = 0,\quad K = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
+$$
 
 Solve the quadratic:
 
-- K = [−9 ± √(9² + 4·10·1)] / (2·10) = [−9 ± √121] / 20 = [−9 ± 11] / 20
-- Feasible (K ≥ 0) root: K = (−9 + 11) / 20 = 2/20 = 0.1
+$$
+K = \frac{-9 \pm \sqrt{9^2 - 4\cdot 10 \cdot (-1)}}{2\cdot 10}
+= \frac{-9 \pm \sqrt{121}}{20}
+= \frac{-9 \pm 11}{20}.
+$$
 
-Therefore, the normalization constant is **K = 0.1**.
+Feasible ($K \ge 0$) root: $\displaystyle K = \frac{2}{20} = 0.1$.
+
+Therefore, the normalization constant is **$K = 0.1$**.