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Probability Disribution Tables

combinations and probablityComputing probabilities involving committeesProbability of selecting different kinds of items from a set of itemsProbability questions.Probability - analyzing “randomness” of dataProbability - HypergeometricProbability Problem (Choosing members in a committee)Precal Probability Expected NumberProbability and how to solve word problemCalculate joint probability interchanging the terms

0

$begingroup$

For the following problem, do I have 9 outcomes or 20? And how do I set up this distribution table (is there a side for students and one for parents?)

Determine the probability distribution, in table form and graphical form, for the number of students selected to be in a six-person fund-raising committee from 9 students and 11 teachers.

probability statistics

share|cite|improve this question

asked May 9 '17 at 0:25

K.AK.A

44

$endgroup$

add a comment | 

0

$begingroup$

For the following problem, do I have 9 outcomes or 20? And how do I set up this distribution table (is there a side for students and one for parents?)

Determine the probability distribution, in table form and graphical form, for the number of students selected to be in a six-person fund-raising committee from 9 students and 11 teachers.

probability statistics

share|cite|improve this question

asked May 9 '17 at 0:25

K.AK.A

44

$endgroup$

add a comment | 

$begingroup$

For the following problem, do I have 9 outcomes or 20? And how do I set up this distribution table (is there a side for students and one for parents?)

Determine the probability distribution, in table form and graphical form, for the number of students selected to be in a six-person fund-raising committee from 9 students and 11 teachers.

probability statistics

share|cite|improve this question

asked May 9 '17 at 0:25

K.AK.A

44

$endgroup$

share|cite|improve this question

asked May 9 '17 at 0:25

K.AK.A

44

share|cite|improve this question

asked May 9 '17 at 0:25

K.AK.A

44

asked May 9 '17 at 0:25

K.AK.A

44

asked May 9 '17 at 0:25

K.AK.A

44

1 Answer
1

active

oldest

votes

1

$begingroup$

You have neither 9 nor 20 outcomes, but 7.

The 7 possible outcomes are having 0,1,2,3,4,5, or 6 students in the committee.

Now, if you have 0 students, you need 6 teachers. Also, there are $9choose0$ ways to pick 0 students out of 9, and $11choose6$ ways to pick 6 teachers out of 11. So, there are $9choose011choose6$ possible committees with 0 students.

Likewise, you get $9choose111choose5$ committees with 1 student, etc.

Finally, to compute probabilities, we divide each of those numbers by the nuber of all possible committees, which is $20choose6$. Of course, that is assuming we pick each of the 20 people with equal probability to be on th committee.

share|cite|improve this answer

edited May 9 '17 at 1:36

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So youre saying on the left column of my table I should go down from 0 to 6, and with the right side would I do 7c0 , 7c1 , and etc?
  $endgroup$
  – K.A
  May 9 '17 at 0:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @K.A No. For $x$ in $0, 1, 2,3,4,5,6$, the probability for selecting $x$ from $9$ students and $6-x$ from $11$ teachers when selecting $6$ from $20$ people is:....
  $endgroup$
  – Graham Kemp
  May 9 '17 at 0:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @K.A Yes, though you also need to choose the appropriate number of teachers to get a full committee of 6 members. So, with 0 students, you get 7C0 * 11C6 possible committees, with 1 student you get 7C1 * 11C5, etc.
  $endgroup$
  – Bram28
  May 9 '17 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bram28 There are 9 students from which to select.
  $endgroup$
  – Graham Kemp
  May 9 '17 at 0:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wait, sorry, but are you sure this is correct? Since it is the probability shouldnt I be dividing and getting a percent?
  $endgroup$
  – K.A
  May 9 '17 at 1:10
  

  
  
  
  

 | 
show 5 more comments

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1

$begingroup$

You have neither 9 nor 20 outcomes, but 7.

The 7 possible outcomes are having 0,1,2,3,4,5, or 6 students in the committee.

Now, if you have 0 students, you need 6 teachers. Also, there are $9choose0$ ways to pick 0 students out of 9, and $11choose6$ ways to pick 6 teachers out of 11. So, there are $9choose011choose6$ possible committees with 0 students.

Likewise, you get $9choose111choose5$ committees with 1 student, etc.

Finally, to compute probabilities, we divide each of those numbers by the nuber of all possible committees, which is $20choose6$. Of course, that is assuming we pick each of the 20 people with equal probability to be on th committee.

share|cite|improve this answer

edited May 9 '17 at 1:36

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So youre saying on the left column of my table I should go down from 0 to 6, and with the right side would I do 7c0 , 7c1 , and etc?
  $endgroup$
  – K.A
  May 9 '17 at 0:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @K.A No. For $x$ in $0, 1, 2,3,4,5,6$, the probability for selecting $x$ from $9$ students and $6-x$ from $11$ teachers when selecting $6$ from $20$ people is:....
  $endgroup$
  – Graham Kemp
  May 9 '17 at 0:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @K.A Yes, though you also need to choose the appropriate number of teachers to get a full committee of 6 members. So, with 0 students, you get 7C0 * 11C6 possible committees, with 1 student you get 7C1 * 11C5, etc.
  $endgroup$
  – Bram28
  May 9 '17 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bram28 There are 9 students from which to select.
  $endgroup$
  – Graham Kemp
  May 9 '17 at 0:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wait, sorry, but are you sure this is correct? Since it is the probability shouldnt I be dividing and getting a percent?
  $endgroup$
  – K.A
  May 9 '17 at 1:10
  

  
  
  
  

 | 
show 5 more comments

1

$begingroup$

You have neither 9 nor 20 outcomes, but 7.

The 7 possible outcomes are having 0,1,2,3,4,5, or 6 students in the committee.

Now, if you have 0 students, you need 6 teachers. Also, there are $9choose0$ ways to pick 0 students out of 9, and $11choose6$ ways to pick 6 teachers out of 11. So, there are $9choose011choose6$ possible committees with 0 students.

Likewise, you get $9choose111choose5$ committees with 1 student, etc.

Finally, to compute probabilities, we divide each of those numbers by the nuber of all possible committees, which is $20choose6$. Of course, that is assuming we pick each of the 20 people with equal probability to be on th committee.

share|cite|improve this answer

edited May 9 '17 at 1:36

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So youre saying on the left column of my table I should go down from 0 to 6, and with the right side would I do 7c0 , 7c1 , and etc?
  $endgroup$
  – K.A
  May 9 '17 at 0:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @K.A No. For $x$ in $0, 1, 2,3,4,5,6$, the probability for selecting $x$ from $9$ students and $6-x$ from $11$ teachers when selecting $6$ from $20$ people is:....
  $endgroup$
  – Graham Kemp
  May 9 '17 at 0:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @K.A Yes, though you also need to choose the appropriate number of teachers to get a full committee of 6 members. So, with 0 students, you get 7C0 * 11C6 possible committees, with 1 student you get 7C1 * 11C5, etc.
  $endgroup$
  – Bram28
  May 9 '17 at 0:58
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Bram28 There are 9 students from which to select.
  $endgroup$
  – Graham Kemp
  May 9 '17 at 0:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Wait, sorry, but are you sure this is correct? Since it is the probability shouldnt I be dividing and getting a percent?
  $endgroup$
  – K.A
  May 9 '17 at 1:10
  

  
  
  
  

 | 
show 5 more comments

$begingroup$

You have neither 9 nor 20 outcomes, but 7.

The 7 possible outcomes are having 0,1,2,3,4,5, or 6 students in the committee.

Now, if you have 0 students, you need 6 teachers. Also, there are $9choose0$ ways to pick 0 students out of 9, and $11choose6$ ways to pick 6 teachers out of 11. So, there are $9choose011choose6$ possible committees with 0 students.

Likewise, you get $9choose111choose5$ committees with 1 student, etc.

Finally, to compute probabilities, we divide each of those numbers by the nuber of all possible committees, which is $20choose6$. Of course, that is assuming we pick each of the 20 people with equal probability to be on th committee.

share|cite|improve this answer

edited May 9 '17 at 1:36

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793

$endgroup$

share|cite|improve this answer

edited May 9 '17 at 1:36

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793

answered May 9 '17 at 0:38

Bram28Bram28

64.2k44793