Problem source

Fire extinguishers may become faulty at any time after  manufacture and are tested annually on the anniversary of  manufacture. The time $T$ years after manufacture until a fire extinguisher becomes faulty is modelled by the continuous probability density function \[
f(t) =
\begin{cases}
\frac{2t}{(1+t^2)^2}& \text{for $t\ge0$}\,,\\[4mm]
 \ \ \ \ 0& \text{otherwise}.
\end{cases}
\]
A faulty fire extinguisher will fail an annual test with probability $p$, in which case it is destroyed immediately. A non-faulty fire extinguisher will always pass the test. All of the annual tests are independent. Show that the probability that a randomly chosen fire extinguisher will be destroyed exactly three years after its manufacture is $p(5p^2-13p +9)/10$. 
Find the probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture.

Solution source

The probability it becomes faulty in each year is:

\begin{align*}
\mathbb{P}(\text{faulty in Y}1) &= \int_0^1 \frac{2t}{(1+t^2)^2} \, dt \\
&= \left [ -\frac{1}{(1+t^2)} \right]_0^1 \\
&= 1 - \frac{1}{2} = \frac{1}{2} \\
\mathbb{P}(\text{faulty in Y}2) &= \frac{1}{2} - \frac{1}{5} = \frac{3}{10} \\
\mathbb{P}(\text{faulty in Y}3) &= \frac{1}{5} - \frac{1}{10} = \frac{1}{10}
\end{align*}

The probability of failing for the first time after exactly $3$ years is:

\begin{align*}
\mathbb{P}(\text{faulty in Y1, }PPF) &+ \mathbb{P}(\text{faulty in Y2, }PF) + + \mathbb{P}(\text{faulty in Y3, }F) \\
&= \frac12 (1-p)^2p + \frac3{10}(1-p)p + \frac1{10}p \\
&= \frac{p}{10} \l 5(1-p)^2 + 3(1-p) + 1 \r \\
&= \frac{p}{10} \l 5 - 10p + 5p^2 + 3 -3p +1 \r \\
&= \frac{p}{10} \l 9 - 13p + 5p^2 \r
\end{align*}

as required.

The probability that a randomly chosen fire extinguisher that was destroyed exactly three years after its manufacture was faulty 18 months after its manufacture is:

\begin{align*}
\mathbb{P}(\text{faulty 18 months after} | \text{fails after 3 tries}) &= \frac{\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})}{\mathbb{P}(\text{fails after exactly 3 tries})}
\end{align*}

We can compute $\mathbb{P}(\text{faulty 18 months after and fails after 3 tries})$ by looking at $2$ cases, fails between $12$ months and $18$ years, and between $0$ years and $1$ year. 

\begin{align*}
\mathbb{P}(\text{faulty between 1y and 18m}) &= \int_{1}^{\frac32} \frac{2t}{(1+t^2)^2} \, dt \\
&= \left [ -\frac{1}{(1+t^2)} \right]_{1}^{\frac32} \\
&= \frac12 - \frac{4}{13} = \frac{5}{26} \\
\end{align*}

So the probability is:

\begin{align*}
\mathbb{P} &= \frac{\frac{5}{26}(1-p)p + \frac12(1-p)^2p}{\frac{p}{10} \l 9 - 13p + 5p^2 \r} \\
&= \frac{\frac{25}{13}(1-p) + 5(1-p)^2}{9 - 13p + 5p^2} \\
&= \frac{5}{13} \frac{(1-p)\l 5 + 13(1-p) \r}{9 - 13p + 5p^2} \\
&= \frac{5}{13} \frac{(1-p)\l 18 - 13p \r}{9 - 13p + 5p^2} \\
\end{align*}