# Uniform Distribution Problem, waiting for plane for 20 min

I have a uniform distribution problem which is as follows: You are waiting for a plane to land. You have reasons to believe that the plane will arrive at a time that is uniformly distributed between 10:30 and 10:50. By 10:37, you’re still waiting for the plane. What is the probability that you will have to wait at least another 5 minutes for the plane?

My first attempt was to integrate from 7 (10:37) to 12 (10:42) for the 5 minutes and I get 25% but that doesn't seem correct. Any advice/suggestions on how to approach this?

Let $X\in[0,\, 20]$ be the uniform random variable representing the number of minutes you should stay after 10:30. What you are asking is $$\Pr(X\ge(5+7)\mid X \ge 7)=\frac{\Pr(X\ge 13)}{\Pr(X\ge 7)}$$
If the arrival was uniformly distributed $[0..20]$ minutes passed 10:30am and has not arrived within 7 minutes, the conditional distribution is uniform on $[7..20]$ minutes passed 10:30am.
So $\mathsf P(X\in(12..20]\mid X\in(7..20])$ is... $\frac{\Box-\Box}{\Box-\Box}$