Fix incorrect solutions for problems in 5.1 and 5.2

docs/Chap05/Problems/5-1.md

@@ -20,4 +20,18 @@ $$
 \frac{n_{i + 1} - n_i}{n_{i + 1} - n_i} = 1.
 $$
 
-**b.** For this choice of $n_i$ , we have that at each increment operation, the probability that we change the value of the counter is $\frac{1}{100}$. Since this is a constant with respect to the current value of the counter $i$, we can view the final result as a binomial distribution with a $p$ value of $0.01$. Since  the variance of a binomial distribution is $np(1 − p)$, and we have that each success is worth $100$ instead, the variance is going to be equal to $0.99n$.
+**b.** For this choice of $n_i$ , we have that at each increment operation, the probability that we change the value of the counter is $\frac{1}{100}$. Since this is a constant with respect to the current value of the counter $i$, we can view the final result as a binomial distribution with a $p$ value of $0.01$. Since the variance of the counter value $C_n$ is given by the formula for a binomial distribution:
+
+$$
+\text{Var}(C_n) = np(1-p) = n \cdot 0.01 \cdot (1 - 0.01) = 0.0099n.
+$$
+
+The question asks for the variance of the value represented by the counter, let's call it $V_n$. The relationship is $V_n = 100 \cdot C_n$.
+
+Using the variance property $\text{Var}(aX) = a^2\text{Var}(X)$, we get:
+
+$$
+\text{Var}(V_n) = \text{Var}(100 \cdot C_n) = 100^2 \cdot \text{Var}(C_n) = 10000 \cdot (0.0099n) = 99n.
+$$
+
+So, the variance is going to be equal to $99n$.

docs/Chap05/Problems/5-2.md

@@ -31,11 +31,10 @@ RANDOM-SEARCH(x, A, n)
     v = Ø                     // a set (or bitmap, etc.) of visited indices
     while |v| < n:
         i = RANDOM(1, n)
-        if i ∉ v:             // only use i if it hasn't been picked before
-            if A[i] = x:
-                return i
-            else:
-                v = v ∪ {i}
+        if A[i] = x:
+            return i
+        if i ∉ v:
+            v = v ∪ {i}
     return NIL
 ```