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  • Logic Five

    A bag has an unknown number of colored objects, with equal numbers of each color. Adding 20 objects of a new color to the bag would not change the probability of drawing (without replacement) two objects of the same color. 

    How many objects are in the bag? (Before the extra objects are added.)


    列出来的等式含有两个未知数,计算的时候有些地方未观察到,导致计算不出结果

    google到了一个解答:点这里 (连接失效,只能用google cache查看)


    假设有 c 种颜色,每种颜色的球 n 个,现在加入一种新颜色的球 k 个,加入前后,不放回的取出2个球同色的概率相同。

    加入前,取出两个球同色的概率:

    Pa = C(1,c) * C(2,n) / C(2,c*n) = (n -1)/(cn - 1);

    加入之后,取出两个球同色的概率:

    Pb = C(1,c) * C(2,n) + C(2,k) / C(2,c*n + k) = [c*n(n-1) + k*(k-1)] / (c*n+k)*(c*n+k-1)

    现在 Pa = Pb,做差

    Pa-Pb = 

     n k (2 n c + k - 1 - c - k c) 
       ---------------------------------
       (n c - 1) (n c + k) (n c + k - 1)

    => nk(2nc + k - 1 -c -kc) = 0   ( n,c,k>0, nc>1)

    => 2nc - (1+k)c + (k-1) = 0  

    显然,k-1 应该是 c 的倍数,设 c=a(k-1) , 等式除以k-1:

    => 2na - (1+k)a + 1 = 0  ……(1)

    注意观察常数项:

    -a + 1 = 0 

    => a = 1       ……(2)

    带入(1)可知:

    n = k/2

    c = k - 1

    具体到本题,k = 20

    所以是19种颜色,每种10个,原先一共有190个球

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  • 原文地址:https://www.cnblogs.com/handt/p/2605573.html
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