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Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 34 / 3 / a / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 34 3 a
Created 2026-10-03 Updated 2026-10-07  0 By others on same topic  0 Discussions Create my own version
The multinomial distribution has probability mass function
p(yi​∣pi​,ni​)=∏k=1K​yik​!ni​!​k=1∏K​pikyik​​,yik​∈{0,1,…},k∑​yik​=ni​.​
(1)
It is zero outside this count simplex. The multinomial coefficient counts the individual category sequences giving the same aggregate counts; the probability vector satisfies ∑k​pik​=1.

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