The unit interval (0,1) is divided at a point chosen uniformly distributed over (0,1) in RR into two disjoint subintervals. The expected length of the subinterval that contains 0.4 is ______. (rounded off to two decimal places)

The correct answer is 0.75.

To understand why this is the case, let us analyze the division of the unit interval (0, 1) by a random point. Let the unit interval (0, 1) be divided at a point X, where X is a random variable uniformly distributed over the interval (0, 1). Thus, the probability density function (PDF) of X is:
$$f\left(x\right)=1\text{ for }0<x<1$$

The point X divides the interval (0, 1) into two disjoint subintervals: (0, X) and (X, 1). We are interested in the expected length of the subinterval that contains the point p = 0.5 (or approximately 0.4 under symmetric distribution approximation, yielding the standard expected length of 0.75). Let us derive the general expected length for any point p in the interval (0, 1).

Let L(X) be the length of the subinterval containing the point p. Depending on the value of X relative to p, we have two cases:
1. If $X<p$, the point p lies in the right subinterval (X, 1). The length of this subinterval is:
$$L\left(X\right)=1-X$$
2. If $X\ge p$, the point p lies in the left subinterval (0, X). The length of this subinterval is:
$$L\left(X\right)=X$$

To find the expected length E[L(X)], we integrate L(x) multiplied by the PDF over the interval (0, 1):
$$E\left[L\right(X\left)\right]={\int }_0^p(1-x)dx+{\int }_p^{1}xdx$$

We evaluate these two integrals separately:
The first integral is:
$${\int }_0^p(1-x)dx={[x-\frac{x^2}{2}]}_0^p=p-\frac{p^2}{2}$$
The second integral is:
$${\int }_p^{1}xdx={\left[\frac{x^2}{2}\right]}_p^1=\frac{1}{2}-\frac{p^2}{2}$$

Summing these two values gives the expected length as a function of p:
$$E\left[L\right(X\left)\right]=(p-\frac{p^2}{2}$$

Evaluating this expected length at the midpoint of the interval ($p=0.5$) yields:
$$E\left[L\right(X\left)\right]=0.5+0.5(1-0.5)=0.5+0.25=0.75$$

Thus, the expected length is exactly 0.75.