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  "body": "\n## Counterexample\n\nToss a fair coin two times, so $\\Omega = \\{HH, HT, TH, TT\\}$ with probability $1/4$ for each outcome.\nLet $A$ be \"the first toss is heads\", $B$ be \"the second toss is heads\", and $C$ be \"the two tosses are equal\".\nThen $A = \\{HH, HT\\}$, $B = \\{HH, TH\\}$ and $C = \\{HH, TT\\}$, so $P(A) = P(B) = P(C) = 1/2$.\nEach intersection of two of them is $\\{HH\\}$, with probability $1/4 = \\frac12 \\cdot \\frac12$. So $A$, $B$ and $C$ are [[pairwise independent|0N57D1XY6T]].\nBut $P(A \\cap B \\cap C) = P(\\{HH\\}) = 1/4$, and $P(A) \\, P(B) \\, P(C) = 1/8$. So $A$, $B$ and $C$ are not [[mutually independent|0GCVEV6BJA]].\n\n## Remarks\n\nHere any two of the events determine the third: if $A$ and $B$ both occur, then $C$ occurs.\nThe same idea with $n$ fair bits and their sums modulo $2$ gives large pairwise independent families. Such families are useful in computer science.\n\n## Explanations\n\n### L1 human\n\nIf you know the first toss, you know nothing about $C$, and the same for the second toss.\nBut if you know both tosses, you know $C$ exactly. So the three events are not independent as a group.\n",
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  "msc": [
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  "name": "Pairwise independent events that are not mutually independent",
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