diff --git a/_problems/unit-02/A-computing-the-distribution-induced-by-a-random-variable/1.md b/_problems/unit-02/A-computing-the-distribution-induced-by-a-random-variable/1.md
index 17cdad7..a220ed2 100644
--- a/_problems/unit-02/A-computing-the-distribution-induced-by-a-random-variable/1.md
+++ b/_problems/unit-02/A-computing-the-distribution-induced-by-a-random-variable/1.md
@@ -5,4 +5,58 @@ statement: |
     $$X(x) = \begin{cases} 1, &x \le 2\\\\ 2, &otherwise\end{cases}$$
     Compute the pmf of $X$
 ---
-Let $Y be a random varable defined with positive probability over the outcome space {1, 2, 3, 4, 5, 6}
+Let $Y$ be a random varable defined with positive probability over the outcome space $\{1, 2, 3, 4, 5, 6\}$
+
+Define the probability of $Y$  as follows:
+
+$$
+P(Y = y) = 
+\begin{cases} 
+\frac{1}{6} & \text{for } y \in \{1,2,3,4,5,6\} \\
+0 & \text{otherwise} 
+\end{cases}
+$$
+
+The implication of the defined transformation is 
+
+$$
+\begin{aligned}
+Y = \{1,2\} & \Rightarrow X = 1 \\
+Y = \{3,4,5,6\} & \Rightarrow X = 2
+\end{aligned}
+$$
+
+In bracket notation, let random variable $X: \Omega \rightarrow \mathbb{R}$ be defined as, 
+
+$$
+X(x) = \begin{cases} 1, &y \le 2\\\\ 2, &y \in \{1,2,3,4,5,6\}\end{cases}
+$$
+
+Computing inducted probabilities, we have
+
+$$
+P(X = 1) = P(Y = 1 \cup Y = 2) = \text{  (Independence)  } P(Y = 1) + P(Y = 2)  = \frac{1}{6} + \frac{1}{6} = \frac{1}{3}
+$$
+
+By law of complementary probability and the definition of the output space of $X$ we have
+
+$$ 
+P(X=2) = 1 - P(X=2) = \frac{2}{3}
+$$
+
+or
+
+$$
+P(X = 2) = P(Y = 3 \cup Y = 4 \cup Y = 5 \cup Y = 6)  = \text{ (Independence) } P(Y = 3) + P(Y = 4) +  P(Y = 5) P(Y = 6) = 
+\frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6}= \frac{2}{3}
+$$
+
+The PMF of X is 
+
+$$
+P(X=x) = \begin{cases} \frac{1}{3}, &x = 1\\\\ \frac{2}{3}, &x =2 \\\\ 0, &otherwise\end{cases}
+$$
+
+
+
+