Draw $x_1, x_2, x_3$ from $U(0, 1)$ what is the probability that the median of $x_1, x_2, x_3$ is $< c$?
The approach that I had in mind is to find the CDF of the second order statistic (the median in this case), differentiate it to get the PDF, and then integrate from 0 to $c$ to get the probability. But is there a more clever approach given that there are only 3 values here?
One approach that I just thought of is to consider the complement, the probability that the median is greater than $c$. This means exactly 2 of $x_1, x_2, x_3$ are $\geq c$ or all 3 are greater than $c$.
The latter occurs with probability $\left(1 - c \right)^3$. The former occurs with probability $\binom{3}{2} \left(1 - c \right)^2 c$.
So the probability that median is less than $c$ should be $$ 1 - \left(1 - c\right)^3 - \binom{3}{2} \left(1 - c\right)^2 c $$ ?
Let the median of $x_1, x_2, x_3$ be denoted $M$. Then $\Pr[M \le c]$ is equivalent to saying that at most one of $x_1, x_2, x_3$ is greater than $c$. This is a binomial probability where the probability that any given $x_i$ is less than or equal to $c$ is equal to $c$; i.e., let $Y \sim \operatorname{Binomial}(c)$ count the number of $x_i$ that are less than or equal to $c$. Then $$\Pr[M \le c] = \Pr[Y \ge 2] = \binom{3}{3} c^3 (1-c)^{3-3} + \binom{3}{2} c^2 (1-c)^{3-2} = c^3 + 3 c^2 (1-c) = c^2(3-2c).$$ This of course is equivalent to your answer.