Probability of the median of three independent outcomes

Question

I have come across a statistics problem that I want to solve but I don't know where to begin. The problem is the following

• What is the probability that the median of three random numbers between 60 and 100 is greater than or equal to 75?

Analytically I was able to use R to calculate that the probability is 0.695884..., however I don't know how mathematically to solve this. Can someone show me how to derive the solution so I can follow the steps and thought-process?

Answer 1

Combine the probability of four cases:

1. $X$, $Y$, and $Z$ are all at least $75$;
2. $X < 75$, but $Y \ge 75$ and $Z \ge 75$;
3. $Y < 75$, but $X \ge 75$ and $Z \ge 75$;
4. $Z < 75$, but $X \ge 75$ and $Z \ge 75$.

These cases are disjoint, and in each of them, the median of $X$, $Y$, and $Z$ is also at least $75$. On the other hand, if none of these cases apply, then at least two of $X$, $Y$, and $Z$ are less than $75$ - making the median less than $75$ as well.

(Also, there is a slight difference in probability here depending on whether we choose $X,Y,Z$ to be random real numbers in the range $[60,100]$, or random integers from the set $\{60,61,\dots,100\}$. Be sure which of these you want!)