(New page: Here's what I did, and it seemed to work. Let me know if I forgot anything. Simply expand the original <math>Pr(x) = \left( \begin{array}{ccc} n \\ x \end{array} \right)p^{x}(1-p)^{n-x}<...) |
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Simply expand the original <math>Pr(x) = \left( \begin{array}{ccc} n \\ x \end{array} \right)p^{x}(1-p)^{n-x}</math>. | Simply expand the original <math>Pr(x) = \left( \begin{array}{ccc} n \\ x \end{array} \right)p^{x}(1-p)^{n-x}</math>. | ||
| − | Now, see if substituting n-x for x results in the same answer. | + | Now, see if substituting n-x for x and expanding results in the same answer. |
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| + | Don't forget that: <math> | ||
| + | {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 </math> | ||
Revision as of 16:38, 23 September 2008
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 $
