Monday

April 21, 2014

April 21, 2014

Posted by **Sarah** on Saturday, September 11, 2010 at 10:29am.

- math -
**drwls**, Saturday, September 11, 2010 at 10:42amThe expected value is sum of n*P(n), for all possible values of n.

0*0.84 + 1*0.13 + 2*0.02 + 3*0.01

= 0.20

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