(New page: We were given an example like this one in class on 9/22/08.<BR><BR> First bar purchased = 1 coupon<BR> X2 = # of extra bars for a different coupon<BR> X2 is geom[(n-1)/n]<BR><BR> X3 = # o...) |
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We were given an example like this one in class on 9/22/08.<BR><BR> | We were given an example like this one in class on 9/22/08.<BR><BR> | ||
| − | First bar purchased = 1 coupon<BR> | + | First bar purchased = 1 coupon<BR><BR> |
| − | + | <math>X_2</math> = # of extra bars for a different coupon<BR> | |
| − | + | <math>X_2</math> is geom<math>(\frac{n-1}{n})</math><BR><BR> | |
| − | + | <math>X_3</math> = # of extra bars after 2nd coupon to get 3rd coupon<BR> | |
| − | + | <math>X_3</math> is geom<math>(\frac{n-2}{n})</math><BR><BR> | |
| − | + | <math>X_4</math> is geom<math>(\frac{n-3}{n})</math><BR><BR> | |
| − | + | <math>X_n</math> is geom<math>(\frac{1}{n})</math><BR><BR> | |
Avg. # of coupons<BR> | Avg. # of coupons<BR> | ||
| − | E[# needed]= | + | E[# needed]=<math>\sum_{i=1}^n E[Xi]</math> |
Latest revision as of 05:28, 5 October 2008
We were given an example like this one in class on 9/22/08.
First bar purchased = 1 coupon
$ X_2 $ = # of extra bars for a different coupon
$ X_2 $ is geom$ (\frac{n-1}{n}) $
$ X_3 $ = # of extra bars after 2nd coupon to get 3rd coupon
$ X_3 $ is geom$ (\frac{n-2}{n}) $
$ X_4 $ is geom$ (\frac{n-3}{n}) $
$ X_n $ is geom$ (\frac{1}{n}) $
Avg. # of coupons
E[# needed]=$ \sum_{i=1}^n E[Xi] $
