自己顶下！

这个第一问，断句是不是有点问题呀。。还有部分地方缩写了。。搞不懂意思
第二问。。两个的时候，用图像法就行了

X可以取1,2,3 Y可以取0,1,2,3  题目很难理解吗？
这种类型的题目出现过啊  一家买了2车保险 一个deductible 1 另一个2 计算总claims少于多少的概率
这里要求Y <=1 且 X>Y  穷举法列出所有可能就行了

第二道也是基本题啊  分N=0, N=1, N=2三种情况啊
P(N=0, Y<=48)=P(N=0)=0.7
P(N=1, Y<=48)=P(N=1) P(Y<=48|N=1) = 0.2 * 48/60 =0.12
P(N=2, Y<=48)=P(N=2) P(Y<=48|N=1) = 0.1 * (0.5* 48 *48)/(60*60)
最好求和就好  
对于均匀分布 一维对应长度 二维对应面积

Ah, X,Y 反了，改了。直接求相对值还是要求和，一样的。
1. P(rebate) = P(Y<=1, X>Y)
    joint pmf f(x,y) = k (8+x-2y) , total probability =1, that is, sum_(x=0)^(3) sum_(y=1)^(3) f(x,y) = 1,
    plug in,  f(x,y) = k (8+x-2y), get:
    k sum_(x=1)^(3) sum_(y=0)^(3) (8+x-2y) = 1,
    notice that:  sum_(x=1)^(3) sum_(y=0)^(3) 8 = 3*4*8=96,  
                          sum_(x=1)^(3) sum_(y=0)^(3) x = 4*sum_(x=1)^(3) x = 4*(1+3)*3/2=24
                          sum_(x=1)^(3) sum_(y=0)^(3) y = 3*sum_(y=0)^(3) y = 3*(0+3)*4/2=18
    therefore, k *(96+24-2*18)=84k =1, thus  k = 1/84  and  the pmf is   f(x,y) = (8+x-2y)/84
    Finally  P(rebate) = P(Y<=1, X>Y) = sum_(y=0,1) sum_(x>y) (8+x-2y)/84
              = (1/84) [sum_(y=0,1) sum_(x>y)8 + sum_(y=0,1) sum_(x>y)x - 2sum_(y=0,1) sum_(x>y)y ]
              = (1/84) { [ sum_(y=0) sum_(x=1,2,3) 8 +sum_(y=1) sum_(x=2,3) 8 ]
                            + [ sum_(y=0) sum_(x=1,2,3) x +sum_(y=1) sum_(x=2,3) x ]
                            - 2 [ sum_(y=0) sum_(x=1,2,3) y +sum_(y=1) sum_(x=2,3) y ] }
              =(1/84) [ 24 + 16 + 6 + 5 - 2*(0+2) ] =47/84


2. let C be the total claim amount,  X be the claim amount when N=1, Y,Z be the claim amounts,  individually, when N=2, then from the question, we know:   X|N=1 ~ Unif(0, 60) ,
   Y,Z|N=2 ~ joint bivariate Uniform on [0,60]x[0,60]
   (and obviously, any other cases have 0 claim amount), so
    P(C<=48) = P(C<=48|N=0)P(N=0) + P(X<=48|N=1)P(N=1) + P(Y+Z<=48|N=2)P(N=2)
= 1*0.7 + (48/60) * 0.2 + [ int_(0<=y<=48) dy int_(0<=z<=48-y) (1/3600) dz } * 0.1

    notice that the double integral is essentially a triangle's area, which is confined by y-axis, z-axis, and the straight line y+z=48, therefore, the double integral is (1/3600)*(1/2)*48*48 = 0.32
    finally,  P(C<=48)=0.7+ 0.8*0.2 + 0.32*0.1 = 0.892

第一题要求是Y<=1 不是X  而且对于这种不需要求k  最后分子分母都有可以约掉  直接计算相对值就可以
X=1,2,3  Y=0,1,2,3  代入数值就可以
#(1,0)= 9  #(1,1)=7 #(1,2)=5 #(1,3)=3 ......
(Y<=1, X>Y)=(1,0) + (2,0) + (2,1) + (3,0) + (3,1)