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  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Introducción al Cálculo de Probabilidades para Estudiantes de Ingeniería\n",
    "#### Código Capítulo 2: Elementos básicos de probabilidad\n",
    "#### `Julia`"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.22"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "a = (300^2 - 100^2)/2\n",
    "p_c = 150^2/(2a)\n",
    "p_b = 1 - 0.5 - p_c"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.32"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "e = [.2,.3,.5];\n",
    "pc₅ = [.20,.24,.10];\n",
    "pc₀ = [.65,.65,.65];`"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "p₅ = sum(pc₅ .* e)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "p₀ = sum(pc₀ .* e)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.33"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "pₑ = .005\n",
    "9e-5*pₑ + 3e-5*(1- pₑ)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.35"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A = [0 .5 .5; .5 0 .5; .5 .5 0]\n",
    "b = [.429; .348; .223]\n",
    "A\\b"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.41"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "A467 = [.95, .82, .92];\n",
    "A18 = [.9, .88];\n",
    "pTCIA = prod(A467)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "pTVA = prod(A18)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "pTCIA + pTVA - pTCIA*pTVA"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.44"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "PS(p)=(1-(1-p)^2)^4;\n",
    "SP(p)=1-(1-p^4)^2;"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "PS(0.6)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "SP(0.6)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.45"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "pUF = 0.97*.04\n",
    "pUFc = 0.025*.96\n",
    "pUF/(pUF+pUFc)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.46"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "RpP = 0.92*0.45;\n",
    "RpPc = 0.12*0.55;\n",
    "VP = RpP/(RpP+RpPc)"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "RnP = 0.88*0.55;\n",
    "RnPc = 0.08*0.45;\n",
    "VF = RnP/(RnP+RnPc)"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Cálculos Ejemplo 2.48"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "B = [.6,.2,.15,.05]\n",
    "MB0 = [.9 .07 .03  0; .05 .88 .05 .02;\n",
    "        .03 .10 .81 .06; .02 .06 .12 .80];\n",
    "m1B = (MB0 .* B);\n",
    "m1 = MB0' * B"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "MB1 = m1B ./ m1';\n",
    "B1 = MB1[:,2]"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "m2B = (MB0 .* B1);\n",
    "m2 = MB0' * B1"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {},
   "outputs": [],
   "source": [
    "MB2 = m2B ./ m2';\n",
    "B2 = MB2[:,2]"
   ]
  }
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