c =0;b=3;a=5;
% PDF
% f(X) = 2*x/15 ; 0 < x < 3;
% f(X) = 1-x/5 ; 3 <= x < 5;
% CDF
% F(X) = x^2/15 ; 0 < x < 3;
% F(X) = x-(x^2+15)/10 ; 3 < x < 5;
x1 = 0:0.1:2.9; x2 = 3:0.1:5;
x = [x1,x2];
F = [1/15*x1.^2,x2-1/10*(x2.^2+15)];
yi = interp1(F, x, rand(1,10000)); % Inverse Interpolation
hist(yi,100)

%Generating Triangular RV for (C=0; B=3; A=5)
C=0; B=3; A=5;
Nu = 10000; Nh = 100;
u = rand(1, Nu); % u ~ RV of Uniform(0,1)
% Evaluate y via IDF
y = (u <= B/(A-C)).*sqrt((A-C)*(B-C)*u) + ...
(u > B/(A-C)).*(A-sqrt((1-u)*(A-C)*(A-B)) ;
hist(y, Nh) % show histogram of y

二楼的附件与三角分布好象无关，楼上的程序代码可以。谢谢

楼主，想请教一下你，

三楼的貌似也不能运行啊：??? y = (u<= B/(A-C)).*sqrt((A-C)*(B-C)*u)+(u>B/(A-C)).*(A-sqrt((1-u)*(A-C)*(A-B)) ;                                         |
Error: Incomplete or misformed expression or statement.

哦，不好意思，楼主楼上的程序是对的，只是倒数第二行最后一句少了一个右括号

So how do you generate random numbers that follow an inverted triangle distribution