Update_10.md

_problems/unit-01/I-law-of-total-probability-and-Bayes-rule/2.md

@@ -38,19 +38,19 @@ First, we find the probability of getting heads $(P(H))$. Since one coin always
 Since each coin is equally likely to be chosen, each has a $1/3$ probability of being chosen. Thus, $P(H)$ is:
 
 \begin{align}
-\[ P(H) = P(C_{hh}) \cdot P(H|C_{hh}) + P(C_{tt}) \cdot P(H|C_{tt}) + P(C_{f}) \cdot P(H|C_{f}) \]
-\[ P(H) = \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot 0.5 \]
-\[ P(H) = \frac{1}{3} + 0 + \frac{1}{6} \]
-\[ P(H) = \frac{1}{2} \]
+\[ P(H) &= P(C_{hh}) \cdot P(H|C_{hh}) + P(C_{tt}) \cdot P(H|C_{tt}) + P(C_{f}) \cdot P(H|C_{f}) \]\\\\
+&= \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 0 + \frac{1}{3} \cdot 0.5 \]\\\\
+&= \frac{1}{3} + 0 + \frac{1}{6} \]\\\\
+&= \frac{1}{2} \]
 \end{align}
 
 Now, using Bayes' Theorem, P(A|H) is:
-
-\[ P(C_{hh}|H) = \frac{P(H|C_{hh}) \cdot P(C_{hh})}{P(H)} \]
-\[ P(C_{hh}|H) = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} \]
-\[ P(C_{hh}|H) = \frac{1}{3} \times 2 \]
-\[ P(C_{hh}|H) = \frac{2}{3} \]
-
+\begin{align}
+\[ P(C_{hh}|H) &= \frac{P(H|C_{hh}) \cdot P(C_{hh})}{P(H)} \]\\\\
+& = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} \]\\\\
+& = \frac{1}{3} \times 2 \]\\\\
+& = \frac{2}{3} \]
+\end{align}
 So, the probability that the coin is the two-headed coin, given that it landed on heads, is 2/3.
 
 **(b) Probability that the same coin will land on heads in the next flip**