182.json

{
  "id": "182",
  "title": "Empirical Probability Mass Function (PMF)",
  "difficulty": "easy",
  "category": "Probability & Statistics",
  "video": "",
  "likes": "0",
  "dislikes": "0",
  "contributor": [
    {
      "profile_link": "https://github.com/jeetmukherjee",
      "name": "jeetmukherjee"
    }
  ],
  "description": "## Problem\n\nGiven a list of integer samples drawn from a discrete distribution, implement a function to compute the empirical Probability Mass Function (PMF). The function should return a list of `(value, probability)` pairs sorted by the value in ascending order. If the input is empty, return an empty list.",
  "learn_section": "\n# Learn Section\n\n# Probability Mass Function (PMF) — Simple Explanation\n\nA **probability mass function (PMF)** describes how probabilities are assigned to the possible outcomes of a **discrete random variable**.\n\n- It tells you the chance of each specific outcome.  \n- Each probability is non-negative.  \n- The total of all probabilities adds up to 1.\n\n## Estimating from data\nIf the true probabilities are unknown, you can estimate them with an **empirical PMF**:\n- Count how often each outcome appears.  \n- Divide by the total number of observations.  \n\n## Example\nObserved sequence: `1, 2, 2, 3, 3, 3` (6 outcomes total)\n- “1” appears once → estimated probability = 1/6  \n- “2” appears twice → estimated probability = 2/6 = 1/3  \n- “3” appears three times → estimated probability = 3/6 = 1/2  \n\n\n    ",
  "starter_code": "def empirical_pmf(samples):\n    \"\"\"\n    Given an iterable of integer samples, return a list of (value, probability)\n    pairs sorted by value ascending.\n    \"\"\"\n    # TODO: Implement the function\n    pass",
  "solution": "from collections import Counter\n\ndef empirical_pmf(samples):\n    \"\"\"\n    Given an iterable of integer samples, return a list of (value, probability)\n    pairs sorted by value ascending.\n    \"\"\"\n    samples = list(samples)\n    if not samples:\n        return []\n    total = len(samples)\n    cnt = Counter(samples)\n    result = [(k, cnt[k] / total) for k in sorted(cnt.keys())]\n    return result",
  "example": {
    "input": "samples = [1, 2, 2, 3, 3, 3]",
    "output": "[(1, 0.16666666666666666), (2, 0.3333333333333333), (3, 0.5)]",
    "reasoning": "Counts are {1:1, 2:2, 3:3} over 6 samples, so probabilities are 1/6, 2/6, and 3/6 respectively, returned sorted by value."
  },
  "test_cases": [
    {
      "test": "print(empirical_pmf([1, 2, 2, 3, 3, 3]))",
      "expected_output": "[(1, 0.16666666666666666), (2, 0.3333333333333333), (3, 0.5)]"
    },
    {
      "test": "print(empirical_pmf([5, 5, 5, 5]))",
      "expected_output": "[(5, 1.0)]"
    },
    {
      "test": "print(empirical_pmf([]))",
      "expected_output": "[]"
    },
    {
      "test": "print(empirical_pmf([0, 0, 1, 1, 1, 2]))",
      "expected_output": "[(0, 0.3333333333333333), (1, 0.5), (2, 0.16666666666666666)]"
    }
  ]
}