Uniform Distribution 2

Suppose we draw three numbers independently from a uniform distribution between 0 and 1, i.e., $\text{Unif}[0, 1]$ . We then sort these numbers in ascending order. What is the probability that the median of these three numbers is less than $\frac{2}{3}$ ?

Answer

Let $$ X_1, X_2, X_3 $$ be the three independent random variables drawn from $\text{Unif}[0, 1]$ . After sorting, let $X_{(1)} \leq X_{(2)} \leq X_{(3)}$ represent the order statistics, where $X_{(2)}$ is the median.

We seek $P(X_{(2)} < \frac{2}{3})$ .

There are two scenarios where the median $X_{(2)}$ is less than $\frac{2}{3}$ :

1. Exactly two numbers are less than $\frac{2}{3}$ , and one is greater than or equal to $\frac{2}{3}$ .

The probability for a single arrangement where two specific numbers are below $\frac{2}{3}$ and one is above is:

\left( \frac{2}{3} \right)^2 \times \left( \frac{1}{3} \right) = \frac{4}{27}

Since there are $\binom{3}{2} = 3$ such arrangements, the total probability for this scenario is:

3 \times \frac{4}{27} = \frac{12}{27}

2. All three numbers are less than $\frac{2}{3}$ .

The probability for this scenario is:

\left( \frac{2}{3} \right)^3 = \frac{8}{27}

Adding both scenarios together, the total probability is:

\frac{12}{27} + \frac{8}{27} = \frac{20}{27}