老师想用问卷的方式调查有多少学生作弊,问卷上只能填yes or no,并且老师给学生一枚硬币。如果学生第一次投硬币得到了Head,那他就根据自己的真实情况填是否作弊,如果是tail,就再投一次,第二次是head的话就一定会填Yes,tail的话就一定填No,最后根据问卷上的yes/No的个数,估算真正作弊的人。.google и
• H -> Y/N: 1/2
• T->T ->N: (1/2)*(1/2)
• T->H ->Y: (1/2)*(1/2)
• total number is N
• total Yes is K-baidu 1point3acres
• Total cheat is X
• cheat rate is p=X/N
• (p*(1/2)+p*(1/4)+(1-p)*(1/4))*N=K
• p*(1/2)+(1/4)=K/N . From 1point 3acres bbs• 2p+1=4K/N
• X=2K-N/2
一共有N个conference room from NO.1 to No.N。有k个meeting独立随机分配到这N个conference room。现在已知1号conference room里面被shcedule了一个meeting,问1号conference room里面被schedule的总共的meeting的数量(已知1号房存在一个meeting,也就是1号房不为空。在这个条件下求1号房总的meeting数的期望。把meeting的集合写成M1,M2,…MK,对任意的i,利用bayes公式计算条件概率)-baidu 1point3acres
• let check all the requirement of binomial distribution are valid:. From 1point 3acres bbs
o Trials are independent (because we can schedule a meeting in any room irrespective of whether they have meetings scheduled or not)
o Fixed number of trials (k)
o P(success) is the same across trials (yes, this is 1/N for every meeting assignment we have to do) . check 1point3acres for more.
• probability of Room 1 is not empty as event A
P(A) = 1-(1-1/N)^k
• probability of meeting i in Rooms 1 as event B
P(B) =1/N. Χ
• conditional probability of meeting i in room 1 is-baidu 1point3acres
P(B | A) = P(A | B) * P(B) / P(A) = 1* P(B) / P(A) = 1/N / (1-(1-1/N)^k)
• the expected number of meeting in room 1 is
n = k * P(B | A) =k/N / (1-(1-1/N)^k)