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  "body": "\n## Proof\n\n1. Let $(\\Omega, \\mathcal{F}, P)$ be a [[probability space|0ZYBZ34RT2]] and let $A, B \\in \\mathcal{F}$ with $0 \u003c P(B) \u003c 1$. Let $B^c = \\Omega \\setminus B$ be the [[complement|0B1FSS1VZW]] of $B$.\n2. By the [[complement rule|0PQEBTYDZM]], $P(B^c) = 1 - P(B)$. Since $P(B) \u003c 1$, we get $P(B^c) \u003e 0$.\n3. The events $B_1 = B$ and $B_2 = B^c$ are disjoint, and $B_1 \\cup B_2 = \\Omega$. Both have positive probability, by the hypothesis and step 2.\n4. The [[law of total probability|0E5G5FMT2R]] for the family $B_1, B_2$ gives $P(A) = P(A \\mid B) \\, P(B) + P(A \\mid B^c) \\, P(B^c)$.\n5. Substituting step 2 gives $P(A) = P(A \\mid B) \\, P(B) + P(A \\mid B^c) \\, (1 - P(B))$.\n\n## Idea\n\nAn event and its complement form the simplest partition of the sample space. The condition $0 \u003c P(B) \u003c 1$ makes both pieces have positive probability, so both conditional probabilities are defined.\n",
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