Conditional Summations

Clash Royale CLAN TAG#URR8PPP

up vote
1
down vote

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1

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.

summation wolfram-alpha-queries

share|improve this question

edited 24 mins ago

asked 1 hour ago

Edward Domka

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
1
down vote

favorite

1

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.

summation wolfram-alpha-queries

share|improve this question

edited 24 mins ago

asked 1 hour ago

Edward Domka

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

up vote
1
down vote

favorite

1

up vote
1
down vote

favorite

1

1

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.

summation wolfram-alpha-queries

share|improve this question

edited 24 mins ago

asked 1 hour ago

Edward Domka

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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.

summation wolfram-alpha-queries

summation wolfram-alpha-queries

share|improve this question

edited 24 mins ago

asked 1 hour ago

Edward Domka

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited 24 mins ago

asked 1 hour ago

Edward Domka

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

share|improve this question

edited 24 mins ago

edited 24 mins ago

edited 24 mins ago

asked 1 hour ago

Edward Domka

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 1 hour ago

Edward Domka

62

asked 1 hour ago

Edward Domka

62

62

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Edward Domka is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

3 Answers
3

active

oldest

votes

up vote
4
down vote

Sum[Binomial[35, n]*100*If[n > 10, 10, n]*(.1^n*.9^(35 - n)), n, 0, 35]

share|improve this answer

answered 1 hour ago

M.R.

15.3k551177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  49 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

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

share|improve this answer

answered 1 hour ago

That Gravity Guy

58637

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
  – Edward Domka
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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).
  – That Gravity Guy
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  25 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

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.

share|improve this answer

answered 13 mins ago

eyorble

4,4401725

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if I am understanding correctly. That reads "minimum n, maximum 10"?
  – Edward Domka
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
  – eyorble
  11 mins ago
  

  
  
  
  

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
4
down vote

Sum[Binomial[35, n]*100*If[n > 10, 10, n]*(.1^n*.9^(35 - n)), n, 0, 35]

share|improve this answer

answered 1 hour ago

M.R.

15.3k551177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  49 mins ago
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

Sum[Binomial[35, n]*100*If[n > 10, 10, n]*(.1^n*.9^(35 - n)), n, 0, 35]

share|improve this answer

answered 1 hour ago

M.R.

15.3k551177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  49 mins ago
  
  

  
  
  
  

add a comment | 

up vote
4
down vote

up vote
4
down vote

Sum[Binomial[35, n]*100*If[n > 10, 10, n]*(.1^n*.9^(35 - n)), n, 0, 35]

share|improve this answer

answered 1 hour ago

M.R.

15.3k551177

Sum[Binomial[35, n]*100*If[n > 10, 10, n]*(.1^n*.9^(35 - n)), n, 0, 35]

share|improve this answer

answered 1 hour ago

M.R.

15.3k551177

share|improve this answer

share|improve this answer

answered 1 hour ago

M.R.

15.3k551177

answered 1 hour ago

M.R.

15.3k551177

answered 1 hour ago

M.R.

15.3k551177

15.3k551177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  49 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  49 mins ago
  
  

  
  
  
  

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
– Edward Domka
49 mins ago

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
– Edward Domka
49 mins ago

add a comment | 

up vote
0
down vote

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

share|improve this answer

answered 1 hour ago

That Gravity Guy

58637

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
  – Edward Domka
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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).
  – That Gravity Guy
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  25 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

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

share|improve this answer

answered 1 hour ago

That Gravity Guy

58637

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
  – Edward Domka
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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).
  – That Gravity Guy
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  25 mins ago
  
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

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

share|improve this answer

answered 1 hour ago

That Gravity Guy

58637

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

share|improve this answer

answered 1 hour ago

That Gravity Guy

58637

share|improve this answer

share|improve this answer

answered 1 hour ago

That Gravity Guy

58637

answered 1 hour ago

That Gravity Guy

58637

answered 1 hour ago

That Gravity Guy

58637

58637

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
  – Edward Domka
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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).
  – That Gravity Guy
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  25 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
  – Edward Domka
  40 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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).
  – That Gravity Guy
  27 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  – Edward Domka
  25 mins ago
  
  

  
  
  
  

I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
– Edward Domka
40 mins ago

I can't get this to work in Wolfram Alpha either. Am i doing something wrong?
– Edward Domka
40 mins ago

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).
– That Gravity Guy
27 mins ago

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).
– That Gravity Guy
27 mins ago

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
– Edward Domka
25 mins ago

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
– Edward Domka
25 mins ago

add a comment | 

up vote
0
down vote

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.

share|improve this answer

answered 13 mins ago

eyorble

4,4401725

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if I am understanding correctly. That reads "minimum n, maximum 10"?
  – Edward Domka
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
  – eyorble
  11 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

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.

share|improve this answer

answered 13 mins ago

eyorble

4,4401725

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if I am understanding correctly. That reads "minimum n, maximum 10"?
  – Edward Domka
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
  – eyorble
  11 mins ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

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.

share|improve this answer

answered 13 mins ago

eyorble

4,4401725

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.

share|improve this answer

answered 13 mins ago

eyorble

4,4401725

share|improve this answer

share|improve this answer

answered 13 mins ago

eyorble

4,4401725

answered 13 mins ago

eyorble

4,4401725

answered 13 mins ago

eyorble

4,4401725

4,4401725

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if I am understanding correctly. That reads "minimum n, maximum 10"?
  – Edward Domka
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
  – eyorble
  11 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  So if I am understanding correctly. That reads "minimum n, maximum 10"?
  – Edward Domka
  11 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
  – eyorble
  11 mins ago
  

  
  
  
  

So if I am understanding correctly. That reads "minimum n, maximum 10"?
– Edward Domka
11 mins ago

So if I am understanding correctly. That reads "minimum n, maximum 10"?
– Edward Domka
11 mins ago

Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
– eyorble
11 mins ago

Min[10,n] is "the smaller of 10 and n". Or, "n, but no more than 10".
– eyorble
11 mins ago

add a comment | 

Edward Domka is a new contributor. Be nice, and check out our Code of Conduct.

 

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