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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2021

Paper 2

After $n$ tosses, the next total is even if either the current total is even and a tail occurs, or the current total is odd and a head occurs. Therefore
$$
q_{n+1}=(1-p)q_n+p(1-q_n)
=p+(1-2p)q_n.
$$
Since $q_0=1$, subtracting the fixed point $1/2$ gives
$$
q_{n+1}-\frac12
=(1-2p)\left(q_n-\frac12\right).
$$
Thus
$$
\boxed{
q_n=\frac{1+(1-2p)^n}{2}}.
$$
The formula also covers the endpoint cases $p=0$ and $p=1$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 3D / Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 3D / Solution

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Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 2 / 3D / Solution

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