1.17是个什么数
问题如下图：
选项：
A. 
B. 
C. 
D. 
解释：
US400,000 US360,000 US370,000 ANSWER: Here, the joint fault probability matters. If the two bon fault, the loss is $1,000,000×(1−60%)+$600,000×(1−40%)=$400,000+$360,000=$760,000\$1,000,000\times\left(1-60\%\right)+\$600,000\times\left(1-40\%\right)=\$400,000+\$360,000=\$760,000$1,000,000×(1−60%)+$600,000×(1−40%)=$400,000+$360,000=$760,000. This will happen with probability 1.27%. The next biggest loss is $400,000, whihprobability of 3.00%-1.27%=1.73% Its cumulative probability must 100.00−1.27=98.73%100.00-1.27=98.73\%100.00−1.27=98.73%. This is slightly above 98%, so $400,000 is the quantile the 98% level of confinor higher. Subtracting the megives $370,000. 请问第二大损失400000发生的概率是如何推导的？
还有最后这个expecteloss (the me) = 40000 是怎么算出来的？
为什么算expecteloss的时候 只考虑回收率而不考虑 fault probability呢？
The next biggest loss is $400,000, whihprobability of 3.00%-1.27%=1.73% 请问为什么概率不是3%*（1-5%）=2.85%
选%的时候，不是应该选择大一点的么？为什么不能选择760的98.7%呢？为什么还需要再算一个400的loss呢？谢谢
