Every minute, a random object might appear on the runway of a certain airport with the probability of $p = \frac {1}{10^6}$. This object will cause a take-off disaster with the probability of $r= \frac{1}{10^3}$. What is the probability that at least one take-off disaster will happen over the coure of a year, given that the state of the runway is controlled every hour?
First of all, I want to find out what the hourly probabiltiy of the crash is. Every object appearing on the runway may or may not cause the crash. The probability that the plane crashes is therefore: $$p(\text{crash})= p(\text{object}) \cdot p(\text{crash | object}) + p(\text{no object}) \cdot p(\text{crash | no object})$$ $$p(\text{crash}) = \frac{1}{10^6} \cdot \frac{1}{10^3} + 0 = \frac{1}{10^9}$$
This is the probabiliy that the plane will crash every minute.
The total probabiliy that the plane will crash over an hour is:
$$p(\text{hour crash}) = 1 - \left( \frac{10^9 - 1}{10^9} \right)^{60} = 6 \cdot 10^{-8} $$
The year has $365 \cdot 24 = 8760$ hours. And so, the probabiliy that the plane does crash is:
$$p(\text{year crash})= 1 - (1 -6 \cdot 10^{-8})^{8760}= 0.00053$$
Do you think my solution is correct?
Your solution is only an approximation (considering yearly probability), as $6\cdot10^{-8} \simeq P($hour crash$)$, however you started in a proper way.
You could ommit calculation of hourly probability and immediately compute yearly probability, as in the year we have $60\cdot24\cdot365=525600$ minutes. And finally the probability
\begin{equation} P(X)=1-\bigg(\frac{10^9-1}{10^9}\bigg)^{525600}. \end{equation}