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An urn contains 6 Red balls and 1 Blue ball. A fair die having faces f1;2;3;4;5;6g is rolled. If the top face on the die shows m, then m random balls are removed from the urn. What is the expected number of Red balls removed by this process?


i have the following in my solutions,

$$\frac{1}{6} \sum_{k=1}^{6} \frac{6}{7}k = \frac{1}{6}\frac{6}{7}\binom{7}{2}$$

can someone please explain where the 7 choose 2 comes from or rather how to get it from the summation?

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1 Answer 1

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$$\sum_{k=1}^nk=\frac{n(n+1)}{2}=\binom{n+1}{2}$$

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  • $\begingroup$ thanks, is there a name that describes this function? $\endgroup$ Commented Mar 17, 2013 at 4:33
  • $\begingroup$ @notamathwiz Yes, I think so, but I can't recall it now. See here for a proof and more. $\endgroup$
    – Julien
    Commented Mar 17, 2013 at 4:35
  • $\begingroup$ @notamathwiz That's the simplest case of Faulhaber's formula. $\endgroup$
    – Julien
    Commented Mar 17, 2013 at 4:38

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