Three numbers are randomly and independently selected from the interval (0, 1). Determine the probability that the median of the three numbers is less than 0.4.
I know that in order for this to happen, we need at least two values to be < 0.4. I'm just not sure where to go from here and what formula to use. Thanks in advance!
On
Comment: Let $X$ be the number of observations below $0.4.$ Then $X \sim \mathsf{Binom}(3, 0.4).$ As you say, the median is below $0.4,$ if $X \ge 2.$
Here are three methods of evaluation, the first of which is @Henry's (+1).
(1) You seek $$P(X \ge 2) = P(X=2)+P(X =3)\\ = 3(.4)^2(.6) + (.4)^3 = 0.352.$$
(2) Also, $P(X \ge 2) = 1 - P(X \le 1) = 0.352;$ computation in R, where pbinom is a binomial CDF.
1 - pbinom(1, 3, .4)
[1] 0.352
(3) Looking at the distribution of the median: Simulation in R of a million iterations: each sampling three independent standard uniform values and taking their median. One can expect three place accuracy.
set.seed(2020)
med = replicate(10^6, median(runif(3)))
mean(med < .4) # aprx. prop. of medians < .4
[1] 0.351944
One can prove analytically that the distribution of the median of three standard uniform random variables $U_i$ is $U_{(2)} \sim \mathsf{Beta}(2,2).$ Here is a histogram of the one million simulated medians along with the density function of this distribution.
hist(med, prob=T, col="skyblue2")
curve(dbeta(x,2,2), add=T, col="orange", lwd=2)
abline(v = .4, col="darkgreen", lwd=2)
Note: The simulation result can be compared with the exact answer as follows.
The density of $\mathsf{Beta}(2,2)$ is $f(x) = 6x(1-x),$ for $ 0<x<1.$ Then $$P(U_{(2)} \le 4) = \int_0^{0.4} 6x(1-x)\,dx = 0.352.$$
diff(pbeta(c(0,.4), 2, 2))
[1] 0.352
Hint: