Past exam of the year 1 of the computer science course of the University of Oxford / 2016 / Discrete mathematics, continuous mathematics and probability (resit) / 7 / b

Put $q=(1-\alpha )(1-\beta )$, the probability of surviving a full day, and $p=1-q=\alpha +\beta -\alpha \beta$. These lie in $(0,1)$. Let $M$ be the event that the first breakdown is in the morning. By independence, surviving $d-1$ complete days and then failing in the morning of day $d$ has probability $q^{d-1}\alpha$.

These events are disjoint, so summing the geometric series givesThere is no missing probability of never breaking down: survival for $d$ days has probability $q^d\to 0$.

Let $J$ count successful journeys before the first breakdown. The probability of surviving the first $r$ journeys is the tail probability $\mathbb{P}(J\ge r)$. For $d\ge 0$, independence giveswhere again $q=(1-\alpha )(1-\beta )$. For a nonnegative integer-valued random variable, the tail-sum formula follows by writing $J={\sum }_{r\ge 1}{1}_{\{J\ge r\}}$ and interchanging nonnegative sums. ThereforeAs a separate first-day check using the law of total expectation, if $E=\mathbb{E}(J)$, then $E=\alpha \cdot 0+(1-\alpha )\beta \cdot 1+q(2+E)$. Solving gives the same finite value. This counts only successful journeys, excluding the failed journey.

Let $D$ be the number of days up to and including the first breakdown, and $M$ the event of a morning first breakdown. For $d=1,2,\dots$, conditional probability givesThusThis geometric distribution counts the day on which the first breakdown occurs. Its parameter is the probability of at least one breakdown during a day, rather than the morning breakdown probability $\alpha$ alone.