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Chebyshev's rule: \[1-\frac{ 1 }{ k ^{2} }\]I've learned this, yes

If you let \(X=5\) then you have \(100\%\) probability that you can serve the customer.

When \(X=4\) you have \(0.19\) probability you turn the customer away.

Seems like no matter what, you have to increase by \(3\) so that you can serve \(5\) at a time.

Why is it when X = 5 would I have an 100% probability of serving the customers?

http://www.wolframalpha.com/input/?i=0.06+%2B0.16+%2B0.26+%2B+0.23+%2B+0.1+%2B+0.19+

Oh wow... Wow. Okay thanks haha. Seemed simple enough.. Thank you!