Update_11.md

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

@@ -55,18 +55,23 @@ So, the probability that the coin is the two-headed coin, given that it landed o
 
 **(b) Probability that the same coin will land on heads in the next flip**
 
+Let's define the events:
+- $H_next$: heads on next flip.
+
 Given that the coin landed heads the first time, there are three possibilities:
 1. It's the two-headed coin, which will always land heads.
 2. It's the fair coin, which has a 50% chance of landing heads.
 3. It's the two-tailed coin, which cannot be the case since we've observed heads.
 
 We've already calculated that the probability of it being the two-headed coin is 2/3, and the probability of it being the fair coin is the remaining 1/3 (since the two-tailed coin is now impossible). Therefore, the probability of the coin landing heads on the next flip is:
 
-\[ P(\text{heads on next flip}) = P(\text{two-headed coin}) \cdot P(\text{heads|two-headed coin}) + P(\text{fair coin}) \cdot P(\text{heads|fair coin}) \]
-\[ P(\text{heads on next flip}) = \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} \]
-\[ P(\text{heads on next flip}) = \frac{2}{3} + \frac{1}{6} \]
-\[ P(\text{heads on next flip}) = \frac{4}{6} + \frac{1}{6} \]
-\[ P(\text{heads on next flip}) = \frac{5}{6} \]
+\begin{align}
+\[ P(H_next) &= P(C_{hh}) \cdot P(H|C_{hh}) + P(C_{f}) \cdot P(H|C_{f}) \]
+&= \frac{2}{3} \cdot 1 + \frac{1}{3} \cdot \frac{1}{2} \]
+&= \frac{2}{3} + \frac{1}{6} \]
+& = \frac{4}{6} + \frac{1}{6} \]
+& = \frac{5}{6} \]
+\end{align}
 
 So, the probability that the same coin will land on heads in the next flip is 5/6