How to calculate the probability that the median exceeds a certain value? Given the pdf $f(x)=\begin{cases}2x&\text{0<x<1}\\0 & \text{otherwise}\end{cases}$. What is the probability that the sample median based on a random sample of size 3 drawn from the distribution with pdf f(x) exceeds 1/2?
Here, although I can calculate the value of the median by integrating f(x) from $0$ to median=m(say) and then equating it to $1/2$. The value for median came $1/\sqrt2$. But I don't know how to find the probability for the median now.
 A: Alexander Pope wrote "A little learning is a dangerous thing, ..." but great learnings (as in BruceET's answer and in the link posted by whuber) can create unneeded diversions in solving little problems.
We have three independent random variables $X_1, X_2, X_3$ for which we readily can compute that $P(X_i < \frac 12) = \frac 14$ and $P(X_i > \frac 12) = \frac 34$. We are asked for the probability that at least two of the $X_i$ exceed $\frac 12$. Well, the probability that all three exceed $\frac 12$ is $\left(\frac 34\right)^3 = \frac{27}{64}$ while the probability that exactly two of the $X_i$ exceed $\frac 34$ is $3\times \frac 14\times \left(\frac 34\right)^2 = \frac{27}{64}$, making the desired probability $\frac{27}{32} = 0.84375$, no muss, no fuss, no Beta distributions or  calculating the pdf of $X_{(2)}$ or simulations in R yielding three digits of accuracy with a million trials.
A: Comment: You have the math in Comments from @Dave and @whuber' link.
The initial probability distribution is a beta distribution, specifically $\mathsf{Beta}(2,1).$ Maybe you can match
your distribution for $H=X_{(2)}$ to results in the following simulation in R, using
a million samples of size $n = 3.$ [With a million iterations you
can expect 2, maybe 3, decimal places of accuracy.]
set.seed(928)
h = replicate(10^6, median(rbeta(3, 2, 1)))
mean(h > .5)
[1] 0.843916       # aprx P(H > .5)
2*sd(h>.5)/10^3
[1] 0.0007258703   # 95% margin of sim err for P(H > .5)
27/32
[1] 0.84375        # exact P(H > .5)   
mean(h)
[1] 0.685836
hist(h, prob=T, br=30, col="skyblue2")
abline(v = .5, lwd=2, col="red")


