Probability Question About Uniform Random Variables and Median Let U, V, W ∼ Uniform(0, 1) be independent. Find the probability that the median
(i.e., the second smallest) of these three random variables lies in the interval (1/4, 3/4).
I cannot figure out what to do. I am also horrible at probability. Thanks for any help. 
 A: Bruce Trumbo's answer is good, but if you want a more basic approach then note that the probability of a single observation  of being between $0.25$ and $0.75$ is $\frac12$, being low is $\frac14$, and being high is $\frac14$.  The middle value will be in the central interval if any of the following patterns happen:


*

*all three observations central, with probability $\frac12 \times \frac12 \times \frac12$

*two central and one low, with probability $3 \times \frac12 \times \frac12 \times \frac14$

*two central and one high, with probability $3 \times \frac12 \times \frac12 \times \frac14$

*one central and one low and one high, with probability $6 \times \frac12 \times \frac14 \times \frac14$


Add these up and you get $\frac{11}{16}=0.6875$.
A: I don't know what you have covered so far, so I won't try to give
you an analytic solution. You may get more help on that if you
would edit your question to give some topics in the chapter preceding the problem, 
speculate on which ones might be related, and say what you think
is keeping you from putting the pieces together.
I will give you simulation results that will give you a very good
clue whether anything you decide to try has provided the right
answer. Simulation shows that the answer is about 0.687.
 m = 10^6;  n = 3;  x = runif(m*n)
 MAT = matrix(x, nrow=m)
 med = apply(MAT, 1, median)
 mean(med > 1/4 & med < 3/4)
 ## 0.687118

Addendum: No other answers and no response yet. So I'll
provide the outline to one analytic approach to an exact solution. Let $X_{(2)}$ denote the
second order statistic (the median) and $F_2(x)$ denote its CDF.
A general formula for order statistics can be simplified
in this instance [three UNIF(0, 1) observations] to the following:
$$F_2(x) = \sum_{i=2}^3 C(3,i) x^i (1 - x)^{n-i},$$
for $0 < x < 1.$ Then $F_2(3/4) - F_2(1/4) = 0.6875,$ which was
approximated by the simulation above about as well as can be expected.
If $U \sim Unif(0,1),$ then $F_U(x) = x$, for $0 < x < 1.$ In the equation above, factor $x^i$ results from $[F_U(x)]^i.$ If the formula used here is not in your textbook you can look on the Internet
for 'order statistics'.
