提问一个统计概率的问题, 求好心人解答

3 workers work on a cluster that has a daily limit of 2 compute ’credits’. When the total daily amount of computing credits of the three data scientists exceeds that daily limit, the company is charged a one thousand dollar penalty. Assume each worker’s daily cluster usage in computing credits is distributed between 0 and 1 and that the chance the usage belongs in some subinterval of [0, 1] is equal to the length of the subinterval. Assume also that the daily usages of the workers across the year are independent. You are asked to estimate the total cost of this penalty over one business year (252 days) with a single number.

我的思路是worker's daily usagexi~Ber(p) 然后p~uniform(0,1)。 一天三个workers' usage~Bin(3,p) 然后P(一天的usage>2| p ~uniform(0,1))。然后我就不会了。。。我觉得应该算个期望啥的但是又好像不对？

我的理解是， sum of independent uniform distribution 是可以计算的，设sum=w，那么在w大于等于2的时候，这个概率密度是f(w)=(w-3)^2/2，也就是说一天当中被惩罚的概率是f(w)从2积分到3为1/6，也就是p(punishment)=1/6，那么p(no punishment)=5/6。所以一天当中罚款的期望是E[punishment]=-1000*1/6+0*5/6=-500/3，然后252天的惩罚的-500/3*252=42000

可能想错了，到时候有答案告诉我下

可以参考https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution，三个相加的时候，在(2,3)的概率密度是1/2(3-x)^2,积分可以得到是1/4.
我不是学这个的仅供参考。

补充内容 (2021-08-28 12:40 +8:00):
好久没积分算错了。。。应该是1/6。

anoymous 发表于 2021-8-27 10:10
. 1point 3acres 可以参考https://en.wikipedia.org/wiki/Irwin%E2%80%93Hall_distribution，三个相加的时候，在(2,3)的概率 ...

should be 1/6

做过这题，是42000，过程楼上有人说了