diff --git "a/\347\254\25401\347\253\240 \347\273\237\350\256\241\345\255\246\344\271\240\346\226\271\346\263\225\346\246\202\350\256\272/1.Introduction_to_statistical_learning_methods.ipynb" "b/\347\254\25401\347\253\240 \347\273\237\350\256\241\345\255\246\344\271\240\346\226\271\346\263\225\346\246\202\350\256\272/1.Introduction_to_statistical_learning_methods.ipynb"
index f66c3a5..7d3df12 100644
--- "a/\347\254\25401\347\253\240 \347\273\237\350\256\241\345\255\246\344\271\240\346\226\271\346\263\225\346\246\202\350\256\272/1.Introduction_to_statistical_learning_methods.ipynb"	
+++ "b/\347\254\25401\347\253\240 \347\273\237\350\256\241\345\255\246\344\271\240\346\226\271\346\263\225\346\246\202\350\256\272/1.Introduction_to_statistical_learning_methods.ipynb"	
@@ -406,7 +406,7 @@
     "**伯努利模型的极大似然估计：**  \n",
     "定义$P(Y=1)$概率为$p$，可得似然函数为：$$L(p)=f_D(y_1,y_2,\\cdots,y_n|\\theta)=\\binom{n}{k}p^k(1-p)^{(n-k)}$$方程两边同时对$p$求导，则：$$\\begin{aligned}\n",
     "0 & = \\binom{n}{k}[kp^{k-1}(1-p)^{(n-k)}-(n-k)p^k(1-p)^{(n-k-1)}]\\\\\n",
-    "& = \\binom{n}{k}[p^{(k-1)}(1-p)^{(n-k-1)}(m-kp)]\n",
+    "& = \\binom{n}{k}[p^{(k-1)}(1-p)^{(n-k-1)}(k-np)]\n",
     "\\end{aligned}$$可解出$p$的值为$p=0,p=1,p=k/n$，显然$\\displaystyle P(Y=1)=p=\\frac{k}{n}$  \n",
     "\n",
     "**伯努利模型的贝叶斯估计：**  \n",