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    "### Data Analysis\n",
    "#### Bayes' rule\n",
    "<hr>\n",
    "https://github.com/ostad-ai/Machine-Learning\n",
    "<br> Explanation: https://www.pinterest.com/HamedShahHosseini/Machine-Learning/background-knowledge"
   ]
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    "The **Bayes' rule** for two events $A$ and $B$:<br><br>\n",
    "$P(A|B)=\\frac{P(B|A) \\cdot P(A)}{P(B)}, P(B) \\neq 0$\n",
    "<br><br> For our example, we measure the probability of a person to be infected if the test is positive<br>\n",
    "<br>$P(Infected|PositiveTest)=\\frac{P(PositiveTest|Infected) \\cdot P(Infected)}{P(PositiveTest)}, P(PositiveTest) \\neq 0$ <br>\n",
    "where $P(PositiveTest)=P(PositiveTest|Infected) \\cdot P(Infected)+P(PositiveTest|not Infected) \\cdot P(not Infected)$"
   ]
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   "source": [
    "def prob_infected_if_test_positive(p_true_positive,p_infected,p_false_positive):\n",
    "    numerator=p_true_positive*p_infected\n",
    "    denominator=numerator+p_false_positive*(1-p_infected)\n",
    "    if denominator!=0:\n",
    "        return numerator/denominator\n",
    "    else:\n",
    "        return None"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "id": "74ae0061",
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     "text": [
      "Probability of person being infected if the test is positive: 0.08411214953271028\n"
     ]
    }
   ],
   "source": [
    "p_true_positive=0.9 # if the person is infected, the test is positive for 90% of cases\n",
    "p_infected=0.02  # the fraction of people in the population to be infected\n",
    "p_false_positive=0.2 # if the person is not infected, the test is positive for 20% of cases\n",
    "# compute the probability of being infected if the test is positive\n",
    "result=prob_infected_if_test_positive(p_true_positive,p_infected,p_false_positive)\n",
    "print(f'Probability of person being infected if the test is positive: {result}')"
   ]
  },
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   "id": "4276ec61",
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   "source": [
    "**Law of total probability:** <br><br>\n",
    "$P(B)=\\sum_i P(B|A_i)P(A_i)$\n",
    "<br><br>**Example:** We have three vases.\n",
    "- vase one has 3 red balls and 6 blue balls.\n",
    "- vase two has 2 red balls and 8 blue balls\n",
    "- vase three has 6 red balls and 3 blue balls\n",
    "\n",
    "We may choose each vase with the equal probability. Then, what is the probability of drawing a red ball from vases?"
   ]
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   "source": [
    "def law_of_total_prob(p_vases=[1/3,1/3,1/3],reds=[3,2,6],blues=[6,8,3]):\n",
    "    p_reds=[]; p_drawn_red=0; p_drawn_blue=0\n",
    "    for red,blue in zip(reds,blues):\n",
    "        p_reds.append(red/(red+blue))\n",
    "    for p_red,p_vase in zip(p_reds,p_vases):\n",
    "        p_drawn_red+=p_red*p_vase\n",
    "        p_drawn_blue+=(1-p_red)*p_vase\n",
    "    return p_drawn_red,p_drawn_blue"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "id": "06446a2d",
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    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Probability of drawing a red ball: 0.39999999999999997\n"
     ]
    }
   ],
   "source": [
    "print(f'Probability of drawing a red ball: {law_of_total_prob()[0]}')"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "id": "4ae0ff9e",
   "metadata": {},
   "outputs": [],
   "source": []
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