{
  "aliases": [
    "chain-rule-of-probability",
    "general-product-rule"
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  "body": "\n## Statement\n\nLet $(\\Omega, \\mathcal{F}, P)$ be a probability space and let $A_1, \\dots, A_n \\in \\mathcal{F}$ with $n \\ge 2$ and $P(A_1 \\cap \\dots \\cap A_{n-1}) \u003e 0$.\nThen\n$$P(A_1 \\cap \\dots \\cap A_n) = P(A_1) \\, P(A_2 \\mid A_1) \\, P(A_3 \\mid A_1 \\cap A_2) \\cdots P(A_n \\mid A_1 \\cap \\dots \\cap A_{n-1}).$$\nHere $P(\\cdot \\mid \\cdot)$ is the [[conditional probability|0FJB06AD2P]].\n\n## Remarks\n\nThe case $n = 2$ is $P(A \\cap B) = P(A) \\, P(B \\mid A)$ for $P(A) \u003e 0$. It is the definition of $P(B \\mid A)$ multiplied by $P(A)$.\nThe hypothesis makes each conditional probability in the formula defined, because each conditioning event contains $A_1 \\cap \\dots \\cap A_{n-1}$.\nExample: draw two cards from a deck without replacement. The probability of two aces is $\\frac{4}{52} \\cdot \\frac{3}{51} = \\frac{1}{221}$.\n\n## Explanations\n\n### L1 human\n\nTo find the chance that several things all happen, take them one at a time.\nMultiply the chance of the first by the chance of the second given the first.\nThen multiply by the chance of the third given the first two, and so on.\n",
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        "url": "https://en.wikipedia.org/wiki/Chain_rule_(probability)"
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