Answer: 0.70:0.80
Let the division point be represented by the random variable X, which is uniformly distributed over the interval (0,1). The probability density function of X is:
f(x)=1for x∈(0,1)
The point X divides the interval (0,1) into two subintervals: (0,X) and (X,1).
The subinterval containing 0.4 depends on the position of X:
- If X≤0.4, the subinterval containing 0.4 is (X,1), which has a length of 1−X.
- If X>0.4, the subinterval containing 0.4 is (0,X), which has a length of X.
Thus, the length of the subinterval containing 0.4, denoted by L(X), is defined piecewise as:
L(X)={1−XXif 0<X≤0.4if 0.4<X<1
The expected length E[L(X)] is calculated as follows:
E[L(X)]=∫01L(x)f(x)dx=∫00.4(1−x)dx+∫0.41xdx
Evaluating the two integrals separately:
∫00.4(1−x)dx=[x−2x2]00.4=0.4−20.16=0.4−0.08=0.32
∫0.41xdx=[2x2]0.41=21−20.16=0.5−0.08=0.42
Adding the two parts together:
E[L(X)]=0.32+0.42=0.74