 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-4/6b/iii/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

{1, 2, \dots}

with success parameter

q = e^{- λ t}

. Its expected value is therefore

E [N (t)] = \frac{1}{q} = e^{λ t} .

Equivalently, multiplying the master equation by

n

and summing gives

m^{'} (t) = λm (t)

with

m (0) = 1

 New to topics? Read the docs here!