# Thread: Committee of Three

1. ## Committee of Three

A committee of three is chosen from five counselors - Adams, Burke, Cobb, Dilby and Evans.
What is the probability Burke is on the committee?

Solution:

3P5 ÷ 3C5

60 ÷ 10 = 6

Here, 6 represents the number of three member groups which include Burke.

Note: 3C5 is the number of committees.

Let A = probability Burke is on the committee.

P(A) = 6/10 = 3/5

Is this correct?

2. ## Re: Committee of Three

Originally Posted by harpazo
A committee of three is chosen from five counselors - Adams, Burke, Cobb, Dilby and Evans.
What is the probability Burke is on the committee?

Solution:

3P5 ÷ 3C5

60 ÷ 10 = 6

Here, 6 represents the number of three member groups which include Burke.

Note: 3C5 is the number of committees.

Let A = probability Burke is on the committee.

P(A) = 6/10 = 3/5

Is this correct?
Permutations have no place in this question.
The answer is $\dfrac{^4\mathscr{C}_2}{^5\mathscr{C}_3}$

3. ## Re: Committee of Three

Originally Posted by harpazo
Note: 3C5 is the number of committees.
No.

4. ## Re: Committee of Three

There are 5 people, any one of which is equally likely to be on the committee. The probability any one of them, and in particular, Burke, is on it is 1/5.

(There are 5 choices for the first person, 4 for the second, 3 for the for the last so there are 5(4)(3)= 60 possible three person ways to form a three person committee from 5 people. If we require that Burke be on the committee, there at 4 choices for the next person and 3 for the last so there are 4(3)= 12 ways to choose a three person committee that includes Burke. The probability that the committee includes Burke is 12/60= 1/5.)

5. ## Re: Committee of Three

Originally Posted by HallsofIvy
There are 5 people, any one of which is equally likely to be on the committee. The probability any one of them, and in particular, Burke, is on it is 1/5.

(There are a total of 5(4)(3)= 60 three person committees, not 10.)
Surely a committee is not an ordered grouping. So that $\dbinom{5}{3}=10$.
Meaning that there are indeed ten different possible committees.
There are $\dbinom{4}{2}=6$ possible committees having Burke as a member.
There are $\dbinom{4}{3}=4$ possible committees not having Burke as a member.

6. ## Re: Committee of Three

Originally Posted by Plato
Permutations have no place in this question.
The answer is $\dfrac{^4\mathscr{C}_2}{^5\mathscr{C}_3}$
Sorry but I do not understand the symbol in the numerator and denominator.

7. ## Re: Committee of Three

Originally Posted by HallsofIvy
There are 5 people, any one of which is equally likely to be on the committee. The probability any one of them, and in particular, Burke, is on it is 1/5.

(There are 5 choices for the first person, 4 for the second, 3 for the for the last so there are 5(4)(3)= 60 possible three person ways to form a three person committee from 5 people. If we require that Burke be on the committee, there at 4 choices for the next person and 3 for the last so there are 4(3)= 12 ways to choose a three person committee that includes Burke. The probability that the committee includes Burke is 12/60= 1/5.)
I finally found the question and answer. The site's answer is 3/5? At first, I also thought the answer to be 1/5.

8. ## Re: Committee of Three

Originally Posted by Plato
Surely a committee is not an ordered grouping. So that $\dbinom{5}{3}=10$.
Meaning that there are indeed ten different possible committees.
There are $\dbinom{4}{2}=6$ possible committees having Burke as a member.
There are $\dbinom{4}{3}=4$ possible committees not having Burke as a member.
The site's answer is 3/5 not 1/5.

9. ## Re: Committee of Three

Originally Posted by harpazo
Sorry but I do not understand the symbol in the numerator and denominator.
Originally Posted by harpazo
The site's answer is 3/5 not 1/5.
There are a number of ways the combinations ate represented. Here are three commonly in use.
$^N\mathscr{C}_k=\mathcal{C}_k^N=\dbinom{N}{k}= \dfrac {N!}{k!(N-k)!}$ that is combination of N items chosen k at a time.

Because $\dfrac{\binom{4}{2}}{\binom{5}{3}}=\dfrac{6}{10}= \dfrac{3}{5}$ the site is correct.

10. ## Re: Committee of Three

Originally Posted by Plato
There are a number of ways the combinations ate represented. Here are three commonly in use.
$^N\mathscr{C}_k=\mathcal{C}_k^N=\dbinom{N}{k}= \dfrac {N!}{k!(N-k)!}$ that is combination of N items chosen k at a time.

Because $\dfrac{\binom{4}{2}}{\binom{5}{3}}=\dfrac{6}{10}= \dfrac{3}{5}$ the site is correct.
Thank you, Plato.