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text: 'Let $\\mathbf H$ be a [random_variable variable] in $\\mathbb P$ for the true hypothesis, and let $H_k$ be the possible values of $\\mathbf H,$ such that $H_k$ is [-1rd]. Then, Bayes' theorem states:\n\n$$\\mathbb P(H_i\\mid e) = \\dfrac{\\mathbb P(e\\mid H_i) \\cdot \\mathbb P(H_i)}{\\sum_k \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)},$$\n\nwith a proof that runs as follows. By the definition of [-1rj],\n\n$$\\mathbb P(H_i\\mid e) = \\dfrac{\\mathbb P(e \\wedge H_i)}{\\mathbb P(e)} = \\dfrac{\\mathbb P(e \\mid H_i) \\cdot \\mathbb P(H_i)}{\\mathbb P(e)}$$\n\nBy the law of [law_of_marginal_probability marginal probability]:\n\n$$\\mathbb P(e) = \\sum_{k} \\mathbb P(e \\wedge H_k)$$\n\nBy the definition of conditional probability again:\n\n$$\\mathbb P(e \\wedge H_k) = \\mathbb P(e\\mid H_k) \\cdot \\mathbb P(H_k)$$\n\nDone.\n\nNote that this proof of Bayes' rule is less general than the [1xr proof] of the [1x5 odds form of Bayes' rule].\n\n## Example\n\nUsing the [22s Diseasitis] example problem, this proof runs as follows:\n\n$$\\begin{array}{c}\n\\mathbb P({sick}\\mid {positive}) = \\dfrac{\\mathbb P({positive} \\wedge {sick})}{\\mathbb P({positive})} \\\\[0.3em]\n= \\dfrac{\\mathbb P({positive} \\wedge {sick})}{\\mathbb P({positive} \\wedge {sick}) + \\mathbb P({positive} \\wedge \\neg {sick})} \\\\[0.3em]\n= \\dfrac{\\mathbb P({positive}\\mid {sick}) \\cdot \\mathbb P({sick})}{(\\mathbb P({positive}\\mid {sick}) \\cdot \\mathbb P({sick})) + (\\mathbb P({positive}\\mid \\neg {sick}) \\cdot \\mathbb P(\\neg {sick}))}\n\\end{array}\n$$\n\nNumerically:\n\n$$3/7 = \\dfrac{0.18}{0.42} = \\dfrac{0.18}{0.18 + 0.24} = \\dfrac{90\\% * 20\\%}{(90\\% * 20\\%) + (30\\% * 80\\%)}$$\n\nUsing red for sick, blue for healthy, and + signs for positive test results, the proof above can be visually depicted as follows:\n\n\n\n%todo: if we replace the other Venn diagram for the proof of Bayes' rule, we should probably update this one too.%',
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