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Thread: Committee of Three

  1. #1
    Super Member harpazo's Avatar
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    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?
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    Re: Committee of Three

    Quote Originally Posted by harpazo View Post
    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 $\dfrac{^4\mathscr{C}_2}{^5\mathscr{C}_3}$
    Thanks from harpazo
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    Re: Committee of Three

    Quote Originally Posted by harpazo View Post
    Note: 3C5 is the number of committees.
    No.
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    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.)
    Last edited by HallsofIvy; Mar 11th 2019 at 01:53 PM.
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    Re: Committee of Three

    Quote Originally Posted by HallsofIvy View Post
    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.
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    Re: Committee of Three

    Quote Originally Posted by Plato View Post
    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.
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  7. #7
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    Re: Committee of Three

    Quote Originally Posted by HallsofIvy View Post
    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.
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    Re: Committee of Three

    Quote Originally Posted by Plato View Post
    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.
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    Re: Committee of Three

    Quote Originally Posted by harpazo View Post
    Sorry but I do not understand the symbol in the numerator and denominator.
    Quote Originally Posted by harpazo View Post
    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.
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    Super Member harpazo's Avatar
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    Re: Committee of Three

    Quote Originally Posted by Plato View Post
    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.
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