Probability

A committe of 5 members will be chosen from a group of 10 teachers and 5 students. What is the probability that the committee
will have
A) all teachers?
B) 3 teachers and 2 students?
C) 3 or 4 teachers?

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  1. You can treat this as a set of Bernoulli Trials and use the binomial formula to find your probability. If we define a 'success' as selecting a teacher and a 'failure' as selecting a student,

    n = number of selections (5)
    r = number of successes
    p = probability of success per trial = 2/3
    q = probability of failure per trial = 1/3

    For these symbols, the probably of 'r' successes is given by:

    P(r) = nCr * p^r * q^(n-r)

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  2. a) There are five successes, r = 5

    P = 5C5 * (2/3)^5 * (1/3)^(5-5)
    = (2/3)^5
    = 32/243

    b) There are three successes, r = 3

    P = 5C3 * (2/3)^3 * (1/3)^(5-3)
    = 10 * (2/3)^3 * (1/3)^2
    = 80/243

    Give the third one a go!

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  3. Binomial distribution requires replacement, constant probability.
    See:
    https://en.wikipedia.org/wiki/Hypergeometric_distribution

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  4. I answered this question yesterday when it was posted under "Angel"
    https://www.jiskha.com/display.cgi?id=1515201345
    Why did you repost it ??

    I stand by those answers for a) and b)
    c) is open for debate since it is ambiguous.

    here is the other way to do a)
    Prob(all 5 are teachers)
    = (10/15)(9/14)(8/13)(7/12)(6/11) = 12/143

    which agrees with my
    C(10,5)/C(15,5) = 252/3003 = 12/143

    b) alternate way:
    a particular case would be TSTTS
    that particular probability of that is
    (10/15)(5/14)(9/13)(8/12)(4/11) = 40/1001

    but those 3 teachers and 2 students can be arranged in
    5!/(3!2!) or 10 ways, so the prob(3 teachers with 2 students) = 10(40/1001) = 400/1001

    which agrees with my
    C(10,3)*C(5,2)/C(15,5) = 400/1001

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