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Past exam of the mathematics course of the University of Oxford

2017

Prelims

Mathematics III

{c}

The event $W>k$ occurs exactly when both $X_1>k$ and $X_2>k$. By <independence>,
$$
\mathbb P(W>k)=q^kq^k=q^{2k}=\left(1-(2p-p^2)\right)^k.
$$
Thus
$$
\boxed{W\sim\operatorname{Geom}(2p-p^2).}
$$
The new success <probability> $2p-p^2$ is the chance that at least one of the two trials succeeds.

Solution

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / iv / Solution

Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / iv / Solution

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Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics III / 4 / a / iv / Solution

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