Revision as of 15:38, 23 September 2008 by Ezarowny (Talk)

Here's what I did, and it seemed to work. Let me know if I forgot anything.

Simply expand the original $ Pr(x) = \left( \begin{array}{ccc} n \\ x \end{array} \right)p^{x}(1-p)^{n-x} $.

Now, see if substituting n-x for x and expanding results in the same answer.


Don't forget that: $ {n \choose k} = \frac{n \cdot (n-1) \cdots (n-k+1)} {k \cdot (k-1) \cdots 1} = \frac{n!}{k!(n-k)!} \quad \mbox{if}\ 0\leq k\leq n \qquad $

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BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman