高一几何概率题任取m∈[-2,5],n∈[-3,1],则关于x的一次函数y＝(m-n)x＋5是增函数的概率为?

Question

高一几何概率题
任取m∈[-2,5],n∈[-3,1],则关于x的一次函数y＝(m-n)x＋5是增函数的概率为?

Answer 1

The answer is 47/56.
The function y is increasing if and only if m>=n.
If m>=1, then m>=n for any n. This contributes (5-1)/(5-(-2))=4/7.
Otherwise -2<=m<=1.
In this case, if -3<=n<=-2, then m>=n for all m. This contributes
(1-(-2))/(5-(-2)) * (-2-(-3))/(1-(-3)) = 3/7 * 1/4.
Otherwise both m and n are in the interval [-2,1]. This contributes
3/7 * 3/4 * 1/2.
Hence the total probability of m>=n is
4/7+3/7*(1/4+3/4*1/2)=47/56.