1) Bernoulli Trial : 2 possible outcome Success = 1 failure = 0 ..\nx is success P(X = x) = p^x (1-p)^1-x ， Mean( expected value) = E[X] = P ， Variance: Var(X) = p(1-p) ，Standard Deviation = 根号 p(1-p) \n2）binomoal distribution ： Mean(expected value) = E[X] = np Variance : Var(X) = np(1-p) Standerdevaitaion = 根号 np(1-p)\n3）Bionomal coefficient: { n k} means number of ways choose k success from n trials.\n4）complementary Probability： 1 - P1) Law of total Probability\n5） Law of total Probability： P(X) = B1 P(X) + B2 (PX) + B3 P(X) = total probaibility randomly\n6）Conditional Probability ： P(A|B) = P( A and B) / P(B) \n7）Bayes’ rule : P(A|B) = P(B|A)P(A)/ P(B)\n8） Central limit / large number\n基础的掌握这些概念 应用到题目中基本就是涉及这些了 下面是一些具体的考试题目， 希望加加米！\nImpression . ----\nX people, 一共有Y Impression随机分配 \n1) expected impression per audience？ . From 1point 3acres bbs\nFor one impression, the prob one person will see = 1/X (Each ad has uniform 1/X prob to be seen). n次实验成功了m次的binomial distribution . Waral dи,\np=1/X, n=Y \nE = np = Y/X \nLet's check all the requirements of binomial distribution are valid: \n1. Trials are independent (because we can assign an impression to any people irrespective of whether they already saw an impression or not) \n2. Fixed number of trials (Y) \n3. P(success) is the same across trials (1/X for every impression assignment we have to do) \n2) probability of each person have at least one impression . 1point 3acres\nFor one user \n- the prob of seeing no impression = (1-1/X)^Y \n- the prob that he/she will see at least 1 ad = 1-(1-1/X)^Y \n~= 1-(1-Y/X) = Y/X \n这里用到了tyler expansion on binomial coefficient: (1 + x)^a ~= 1 + ax 需要X很大 \n3) expected number of audience have at least one impression -baidu 1point3acres\nthe number of people who will see at least one impression = X (1-(1-Y/X)^Y) 插入广告 \n是关于脸书的广告。假如有两种不同方法来往信息流里插入广告。 \n第一种是：每个信息位子，我们以4%的概率来替换成一个广告。 \n第二种是：每25个信息，把其中一个变成广告。 \nQ1: For each option, what is the expected number of ads shown in 100 news stories? Variance? \n1) binomial distribution: p=0.04, n=100, E=4, V=3.84 \n2) E=4, V=0 \nQ2: Probability of seeing over than twice of expected value \n1) 1-p(x<=8)\nnormal approximation: P(x>8) = P(z > (8-4)/3.84) = 1-pnorm(1.04) \n2) 0 . Χ\nQ3: Max number of back to back ads, which one is more likely \n1) Expected value of back-to-back ads \nexpected number of ads collision, 就是连续两个post都是ads。等于是有了个新的Bernoulli process，概率p=0.04^2. 考虑collision是否independent，写了个模拟发现自己没算错，相邻 两个点是否出现collision是同时决定的，所以independent \n99(0.04^2) .--\n3(0.04^2) \n2) Max number of back-to-back ads \n99 vs 2 \n3) Probability of seeing back-to-back ads .\n- 100个里面有k个广告，k>50则有连续。算k>50个广告的概率 \n插空法：把100-k个广告插入k个广告中间 \n求和k from 0 to 50（100-k+2 选 k） 0.04^k 0.96^(100-k) \n- 3(0.04^2) - (0.04^2)^2 \nConference Room \n1) 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schedule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meet