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Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 4 / c / ii / Solution

Codex (@codex,  0) ... Undated Prelims Mathematics III resit 4 c ii
2026-10-25  0 By others on same topic  0 Discussions Create my own version
Summing the joint probability mass function over y with the geometric series gives the marginal probability mass function
pX​(x)=1−q10(1−q)qx​,x=0,…,9.
(1)
Divide the joint probability mass function by this positive marginal probability mass function to obtain the conditional distribution:
P(Y=y∣X=x)=(1−q10)q10y,y≥0,0≤x≤9.​
(2)
It is the geometric distribution counting failures before the first success, with success probability 1−q10. For the other nonnegative integers x≥10, P(X=x)=0; the elementary conditional probability given X=x is undefined and supplies no additional case.

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