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Mar 18, 2021 at 8:52 comment added Henry @user71207 You can say ${n \choose k}(\frac{1}{2})^k (\frac{1}{2})^{n-k} = {n \choose n-k}(\frac{1}{2})^{n-k} (\frac{1}{2})^{k}$
Mar 18, 2021 at 6:36 comment added user71207 Can you explain your first line? $(\frac{1}{2})^k ≠ (\frac{1}{2})^{n-k}$
Nov 18, 2011 at 22:02 vote accept problab
Nov 18, 2011 at 21:58 history answered Henry CC BY-SA 3.0