0 like 0 dislike
0 like 0 dislike
Probability Question

1 Answer

0 like 0 dislike
0 like 0 dislike
First of all we can assume k >= m since otherwise we have a 100% chance that one server fails.

Next how many options do the first job have? Its k. The second job have k-1 until the mth job has k-m+1 possibilities.

So this yields k!/(k-m)! Diffrent possibilieties to assign the jobs without fail.

In total we have k^m possibilieties to asign the jobs in total.

This yield a probability of k!/((k-m)!k^(m))

No related questions found

24.8k questions

103k answers


33.7k users

OhhAskMe is a math solving hub where high school and university students ask and answer loads of math questions, discuss the latest in math, and share their knowledge. It’s 100% free!