企业官网建设流程全解析

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-i+1)/NXi服从几何分布，E[Xi]=N/(N-i+1)
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：N=5时，E≈5×2.28=11.4盒N=50时，E≈50×4.50=225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-Y<=5，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=3600-3025=575答案：P=575/3600=23/144≈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