概率题

1. 《编程之美》读书笔记13： 4.1 金刚坐飞机问题
   
   问题：现在有一班飞机将要起飞，乘客们正准备按机票号码（1, 2, 3, …N）依次排队登机。突然来了一只大猩猩（对，他叫金刚）。他也有飞机票，但是他插队第一个登上了飞机，然后随意地选了一个座位坐下了1。根据社会的和谐程度，其他的乘客有两种反应：
   1. 乘客们都义愤填膺，“既然金刚同志不遵守规定，为什么我要遵守？”他们也随意地找位置坐下，并且坚决不让座给其他乘客。
   2. 乘客们虽然感到愤怒，但还是以“和谐”为重，如果自己的位置没有被占领，就赶紧坐下，如果自己的位置已经被别人（或者金刚同志）占了，就随机地选择另一个位置坐下，并开始闭目养神，不再挪动位置。
   那么，在这两种情况下，第 i 个乘客（除去金刚同志之外）坐到自己原机票位置的概率分别是多少？
   
   对问题一，每个人都是随机选择座位，任意一个人坐在指定座位的概率相同，因而第i个乘客坐在其座位的概率是 1/n。
   对问题二，答案和金刚的原来座位编号有关。不妨先去除金刚的座位，将乘客（根据机票号）和剩下的座位，按原大小顺序从1开始重新编号。用F(i，n)表示在新排列中（共有n-1个乘客座位和金刚原来的座位），新的第i个乘客坐在其原来座位的概率，则在n个座位中：
   
   ① 金刚若挑自己的座位或选的座位在第i个座位后（共有n-i个座位满足这个条件），则第i个乘客肯定能坐到原来的座位；
   
   ② 金刚若挑选的座位在第i个座位前，不妨假设为j，则第j个乘客除非坐到金刚的座位，不然就会抢其他人的座位，因为他的行为和金刚相似，可以将他当做金刚处理。去除前j个座位，剩下的座位和乘客再按原大小排序重新从1开始编号，则先前的第i个乘客，其座位号变为i-j，新的总座位数变为n-j。因而可得公式：
   
   用 G(i, n)表示原排列中，第i个乘客坐到自己座位的概率，假设金刚的座位编号为j。
   
   若 i<j 则   G(i,n)=F(i,n)= (n-i)/(n+1-i)。
   
   若 i>j 则   G(i,n)=F(i-1,n)= (n+1-i)/(n+2-i) 。
    
   
2. Boys and Girls     (http://toostep.com/insight/google-interview-questions-part-1)
   In a country where everyone wants a boy, each family continues having babies till they have a boy. After some time, what is the proportion of boys to girls in the country? (Assuming probability of having a boy or a girl is the same)
   
   Answer:-
   
   This is a very simple probability question in a software interview. This question might be a little old to be ever asked again but it is a good warm up.
   
   Assume there are C number of couples so there would be C boys. The number of girls can be calculated by the following method.
   
   Number of girls = 0*(Probability of 0 girls) + 1*(Probability of 1 girl) + 2*(Probability of 2 girls) + …
   Number of girls = 0*(C*1/2) + 1*(C*1/2*1/2) + 2*(C*1/2*1/2*1/2) + …
   Number of girls = 0 + C/4 + 2*C/8 + 3*C/16 + …
   Number of girls = C
   (using mathematical formulas; it becomes apparent if you just sum up the first 4-5 terms)
   
   The proportion of boys to girls is 1 : 1.
   
   
   
3. Probability of a Car Passing By        (http://toostep.com/insight/google-interview-questions-part-1)     
   
   The probability of a car passing a certain intersection in a 20 minute windows is 0.9. What is the probability of a car passing the intersection in a 5 minute window? (Assuming a constant probability throughout)
   
   
   Answer:-
   
   This is one of the basic probability question asked in a software interview. Let’s start by creating an equation. Let x be the probability of a car passing the intersection in a 5 minute window.
   
   Probability of a car passing in a 20 minute window = 1 – (probability of no car passing in a 20 minute window)
   Probability of a car passing in a 20 minute window = 1 – (1 – probability of a car passing in a 5 minute window)^4
   0.9 = 1 – (1 – x)^4
   (1 – x)^4 = 0.1
   1 – x = 10^(-0.25)
   x = 1 – 10^(-0.25) = 0.4377