more fixed typo in Maximum_likelihood.ipynb

Author: unpingco

Maximum_likelihood.ipynb

@@ -344,11 +344,11 @@
       "\n",
       "Let's plug in some live numbers here for our worst case scenario where $p=1/2$. Then, if $\\epsilon = 1/100$, we have\n",
       "\n",
-      "$$ \\mathbb{P}\\left( \\frac{99 n}{200} \\lt \\sum_{i=1}^n x_i   \\lt \\frac{101 n}{100} \\right)$$\n",
+      "$$ \\mathbb{P}\\left( \\frac{99 n}{200} \\lt \\sum_{i=1}^n x_i   \\lt \\frac{101 n}{200} \\right)$$\n",
       "\n",
       "Since the sum in integer-valued, we need $n> 100$ to even compute this. Thus, if $n=101$ we have\n",
       "\n",
-      "$$ \\mathbb{P}\\left( \\frac{9999}{100} \\lt \\sum_{i=1}^{101} x_i   \\lt \\frac{10201}{100} \\right) = f\\left(\\sum_{i=1}^{101} x_i  = 50,p\\right)= \\binom{101}{50} (1/2)^{50}  (1-1/2)^{101-50} = 0.079$$\n",
+      "$$ \\mathbb{P}\\left( \\frac{9999}{200} \\lt \\sum_{i=1}^{101} x_i   \\lt \\frac{10201}{200} \\right) = f\\left(\\sum_{i=1}^{101} x_i  = 50,p\\right)= \\binom{101}{50} (1/2)^{50}  (1-1/2)^{101-50} = 0.079$$\n",
       "\n",
       "This means that in the worst-case scenario for $p=1/2$, given $n=101$ trials, we will only get within 1% of the actual $p=1/2$ about 8% of the time. If you feel disappointed, that only means you've been paying attention. What if the coin was really heavy and it was costly to repeat this 101 times? Then, we would be within 1% of the actual value only 8% of the time. Those odds are terrible.\n",
       "\n",