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I am not a math major, but for a networking class, I am taking I am required to do summations for probability. I know the logic but I don't know the mathematical theory to make this work. Using Wolfram I'm trying to do this

sum [(35 choose n)* (100*j)*(.1^n* .9^(35-n))] from n=0 to 35, if[n>10, j=10, j=n]

Essentially, what I'm asking is: How do I run a summation with a variable that is affected by an if statement in this way? Or is there a different math I should be using.

Edit: Sorry for not being clear. I am trying to do this in Wolfram Alpha. I did not realize there was a difference.

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    $\begingroup$ I'm voting to close this question as off-topic because questions concerning Wolfram|Alpha are specifically excluded from this site $\endgroup$
    – m_goldberg
    Commented Sep 18, 2018 at 20:32
  • $\begingroup$ Okay, I am sorry. I did not realize $\endgroup$
    – Edward Domka
    Commented Sep 18, 2018 at 20:35

4 Answers 4

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Sum[Binomial[35, n]*100*If[n > 10, 10, n]*(.1^n*.9^(35 - n)), {n, 0, 35}]
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  • $\begingroup$ Thank you so much! I'm used to programming, I couldn't fully grasp how the if statements worked. I tried looking here but was confused by all the high level stuff. This is perfect. Thank you so much edit: Wolfram didn't understand $\endgroup$
    – Edward Domka
    Commented Sep 18, 2018 at 18:33
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Sum[Binomial[35,n]100*Min[10,n]*.1^n*.9^(35-n), {n,0,35}]

349.946

Nested If statements work well in Mathematica, but Wolfram|Alpha seems to have some issues with them. However, in this case using Min to grab the minimum of n and 10 achieves the appropriate behavior.

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  • $\begingroup$ So if I am understanding correctly. That reads "minimum n, maximum 10"? $\endgroup$
    – Edward Domka
    Commented Sep 18, 2018 at 19:10
  • $\begingroup$ Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10". $\endgroup$
    – eyorble
    Commented Sep 18, 2018 at 19:11
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You can also get Expectation of transformed random variable y = 100 Min[10, x] where x is a random variable with distribution BinomialDistribution[35, .1]:

Expectation[100 Min[10, x], Distributed[x, BinomialDistribution[35, .1]]]

349.94631523961846

Alternatively,

td = TransformedDistribution[100 Min[10, x], Distributed[x,BinomialDistribution[35, .1]]];
Expectation[y, Distributed[y, td]]

349.94631523961846

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Another way

Sum[Binomial[35,n]100 j .1^n .9^(35-n)/.{j:>10/;n>10,j:>n/;n<=10},{n,0,35}]

349.946

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  • $\begingroup$ I can't get this to work in Wolfram Alpha either. Am i doing something wrong? $\endgroup$
    – Edward Domka
    Commented Sep 18, 2018 at 18:41
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    $\begingroup$ If you are trying to evaluate this sum using Wolfram|Alpha and not Wolfram Mathematica, I would edit your post to make that explicitly clear (though I can see where the confusion comes from). $\endgroup$
    – NonDairyNeutrino
    Commented Sep 18, 2018 at 18:55
  • $\begingroup$ Okay, I will. Sorry, I thought they used the same code. That's why I tagged it Wolfram-alpha-queries. I'm new to this $\endgroup$
    – Edward Domka
    Commented Sep 18, 2018 at 18:56

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