A car is new at the beginning of a calendar year. The time, in years,
before the car experiences its first failure is exponentially
distributed with mean 2. Calculate the probability that the car
experiences its first failure in the last quarter of some calendar
year.
Attempt
Let $T$ be rv in years before the car experiences its first failure. We know $T$ is $exp(\lambda = 1/2) $. We want to find
$$ P( 1 > T > 3/4) = F(1) – F(3/4) = e^{-3/4} – e^{-1/2} \approx 0.0808$$
but the answer in the book of my books gives $\boxed{0.205}$. What is my mistake?
Best Answer
I think you just misread the question, it means the first failure in the last quarter of some calendar year, it need not be the first year.
\begin{align} \sum_{n=0}^\infty P\left(n+1 > T > n + \frac34\right) &=\sum_{n=0}^\infty \left(F(n+1)-F\left(n+\frac34\right)\right)\\ &= \sum_{n=0}^\infty \left(\exp \left(-\frac{n+\frac34}2 \right) - \exp\left( -\frac{n+1}{2}\right) \right)\\ &= \left( \exp\left(-\frac38 \right) -\exp\left( -\frac12\right)\right)\sum_{n=0}^\infty \exp(-n/2) \\ &=\frac{\left( \exp\left(-\frac38 \right) -\exp\left( -\frac12\right)\right)}{1-\exp(-1/2)} \\ &\approx 0.205 \end{align}