The factors of 10 are 5 and 2. Note that the 5th multiple of 5 will bring a pair of 5s into the product, which means you couldn t group just 5 5s as the factorials of 25+ have 6 fives within it. Every even number contributes a 2 so that is where there are plenty of 2s to perform a sort of re-arranging of the prime factorization to get 10^6 within all factorials of 25 or higher. Under 25, there are at most 4 factors that contain a 5: 5, 10, 15, and 20.
Given an array of size n I want to generate random probabilities for each index such that Sigma(a[0]..a[n-1])=1 One possible result might be: 0 1 2 3 4 0.15 0.2 0.18 0.22 0.25 ...
Selecting without any weights (equal probabilities) is beautifully described here. I was wondering if there is a way to convert this approach to a weighted one. I am also interested in other ...
10个帐篷的清单有10个。 可能的命令或变更。 为什么是任意的。 只剩下5 000个工件后,便可复制件?
This problem is a little similar to that solved by reservoir sampling, but not the same. I think its also a rather interesting problem. I have a large dataset (typically hundreds of millions of ...
I need to calculate the probability mass function, and cumulative distribution function, of the binomial distribution. I would like to use MATLAB to do this (raw MATLAB, no toolboxes). I can calculate ...
We ve all poked fun at the X minutes remaining dialog which seems to be too simplistic, but how can we improve it? Effectively, the input is the set of download speeds up to the current time, and ...
for(i=1;i<=n;i++) { pick a random index j between 1 and n inclusive; swap card[i] and card[j]; } for the above code am trying to find the probability of original card[k] winding up in slot ...
Is there a direct way of getting the Nth combination of an ordered set of all combinations of nCr? Example: I have four elements: [6, 4, 2, 1]. All the possible combinations by taking three at a time ...
