8 pages read

study/001_introduction-to-probability-statistics-and-random-processes/README.md

@@ -2,4 +2,4 @@
 
 total pages=1007
 
-**Currently reading:** chapter 1, page 80
+**Currently reading:** chapter 1, page 88

study/001_introduction-to-probability-statistics-and-random-processes/ch1/notes.ipynb

@@ -685,6 +685,60 @@
     "$$p_1 = \\frac{p_2}{1 + p_2}$$"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "7be1e873",
+   "metadata": {},
+   "source": [
+    "### Example 1.24\n",
+    "\n",
+    "Let $A$ be the event where my chosen marble is red.\n",
+    "\n",
+    "Let $B_1$ be the set of the first bag, $B_2$ the second bag, $B_3$ the third bag.\n",
+    "\n",
+    "Notice that $B_1, B_2,$ and $B_3$ are disjoint and therefore form a partition of our sample space.\n",
+    "\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(A) &= P(A|B_1)P(B_1) + P(A|B_2)P(B_2) + P(A|B_3)P(B_3) \\\\\n",
+    "&= \\frac{75}{100} \\cdot \\frac{1}{3} + \\frac{60}{100} \\cdot \\frac{1}{3} + \\frac{45}{100} \\cdot \\frac{1}{3} \\\\\n",
+    "&= \\frac{180}{300} = \\frac{6}{10} = 0.6\n",
+    "\\end{align*}\n",
+    "\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "a3160c4c",
+   "metadata": {},
+   "source": [
+    "### Example 1.25\n",
+    "\n",
+    "Let $A$ be the event we chose a red marble and let $B$ be the event we chose from the first bag.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(B|A) &= \\frac{P(A|B)P(B)}{P(A)} \\\\\n",
+    "&= \\frac{\\frac{75}{100} \\cdot \\frac{1}{3}}{\\frac{6}{10}} \\\\\n",
+    "&= \\frac{75}{300} \\cdot \\frac{10}{6} \\\\\n",
+    "&= \\frac{750}{1800} = \\frac{75}{180} = \\frac{15}{36} = \\frac{5}{12} \\\\\n",
+    "\\end{align*}\n"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "34782cd8",
+   "metadata": {},
+   "source": [
+    "### Example 1.26\n",
+    "\n",
+    "Let $A$ be the event we chose a red marble and let $B$ be the event we chose from the first bag.\n",
+    "\n",
+    "\\begin{align*}\n",
+    "P(B|A) &= \\frac{P(A|B)P(B)}{P(A)} \\\\\n",
+    "\n",
+    "\\end{align*}"
+   ]
+  },
   {
    "cell_type": "code",
    "execution_count": null,

study/001_introduction-to-probability-statistics-and-random-processes/ch1/summary.ipynb

@@ -133,13 +133,44 @@
     "$$P(A_1 \\cup A_2 \\cup \\dots \\cup A_n) = 1 - (1 - P(A_1))(1 - P(A_2))\\dots(1 - P(A_n))$$"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "90cf5b00",
+   "metadata": {},
+   "source": [
+    "## Law of Total Probability (1.4.2)\n",
+    "\n",
+    "$$P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)$$\n",
+    "\n",
+    "**Definition.** Law of Total Probability: \\\n",
+    "If $B_1, B_2, B_3, \\dots $ is a partition of the sample space $S$, then for any event $A$ we have\n",
+    "\n",
+    "$$P(A) = \\sum_i P(A \\cap B_i) = \\sum_i P(A|B_i)P(B_i)$$"
+   ]
+  },
   {
    "cell_type": "markdown",
    "id": "c33df800",
    "metadata": {},
    "source": [
-    "\n"
+    "## Bayes' Rule (1.4.3)\n",
+    "\n",
+    "**Definition.** Bayes' Rule\n",
+    "\n",
+    "- For any two events $A$ and $B$, where $P(A) \\neq 0$, we have\n",
+    "\n",
+    "$$P(B|A) = \\frac{P(A|B)P(B)}{P(A)}$$\n",
+    "\n",
+    "- If $B_1, B_2, B_3, \\dots$ form a partition of the sample space $S$, and $A$ is any event with $P(A) \\neq 0$, we have\n",
+    "\n",
+    "$$P(B_j|A) = \\frac{P(A|B_j)P(B_j)}{\\sum_i P(A|B_i)P(B_i)}$$"
    ]
+  },
+  {
+   "cell_type": "markdown",
+   "id": "849d7c99",
+   "metadata": {},
+   "source": []
   }
  ],
  "metadata": {