原题(绿皮书Ch4Probability)There are N distinct types of coupons in cereal boxes and each type, independent of prior selections, is equally likely to be in a box. If a chid wants to collect a complete set of coupons with at east one of each ype, how many coupons (boxes) on average are needed o make such a complete set?中文题意麦片盒里随机附赠N种卡片中的一种(等概率独)。要集齐所有N种卡片平均需要买多少盒?解答定义Xi为已有i-1种后收集到第i种新卡片所需的额外盒数。已有i-1种时下一盒是新品的概率p (N-i1)/NXi服从几何分布E[Xi]N/(N-i1)总期望: E[X] E[X1] E[X2] … E[XN] N/N N/(N-1) N/(N-2) … N/1 N × (1 1/2 1/3 … 1/N) N × H(N)其中H(N)是第N个调和数约等于In(N)0.577欧拉常数)。例N5时E≈5×2.2811.4盒N50时E≈50×4.50225盒答案N×H(N)≈N·In(N)#量化面试#Quant #期望值#几何分布#调和级数原题(绿皮书Ch4Probability)Two bankers each arrive at the station at some random time between 5:00 am and 6:00 am (arrival time for either banker is uniformlydistributed). They stay exactly five minutes and then leave. What is the probability they wil meet on a given day?中文题意两位银行家分别在5:00~6:00之间的某个随机时刻到达车站(均匀分布相互独立)。每人只停留5分钟就离开。他们相遇的概率是多少解答建模设A在第X分钟到达B在第Y分钟到达。X,Y~Uniform(0,60)独立。相遇条件X-Y|5转化为几何概率样本空间60×60正方形面积3600有利区域X-Y5即对角线附近宽5的带状区域不相遇区域两个角落的直角三角形边长55不相遇面积2×(1/2×55²)3025相遇面积3600-3025575答案P575/360023/144≈15.97%通用公式时间窗口T分钟停留t分钟P(相遇)1-(T-t)/T)^2#量化面试#Quant#几何概率#连续分布#面积法