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发帖数: 180 | 1 1138个连号申请者中，随机抽取514，其中有任意14个(及以上)是连号的可能性有多大?
经华中科技大学概率统计系副主任王湘君计算，此次14连号的概率约为百分之一。想
了半天想不出算法该如何算？
之前武汉5141人抽取124名市民中，有6人是连号。记者采访数学博士，说几率是千万亿
分之一。有这么小吗?
有哪位高人指点一下? | p*****k
发帖数: 318 | 2 it's straightforward to prove that the number of ways, in which N integers are chosen from a total of M consecutive integers, so that no length of at least k consecutive integer sequence appears, is the coefficient of the x^N term in the expansion of
[1+x+x^2+...+x^(k-1)]^(1+M-N)=(1-x^k)^(1+M-N)*(1-x)^[N-M-1]
hence the appearance of at least k consecutive integer sequence has the prob. of
1-sum{from i=0 to floor(N/k)} (-1)^i*C(1+M-N,i)*C(M-k*i,M-N)/C(M,N)
with M=1138, N=514, k=14, mathematica gi |
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