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in a large survey, it was discovered that 1 out of every 5 adults visits disney world every year. if 30 adults are randomly selected, what is the probability that exactly 7 of them will visit disney world this year.

  • math -

    prob of visit = 1/5 = p = .2
    Binomial dist:
    p(x = 7) = [30:7] .2^7 (.8)^23
    look up binomial coef [30:7]
    or use formula
    30!/[7!(23!)]

    30!/23!/7! = 30*29*28*27*26*25*24/(7*6*5*4*3*2)
    = 29*28*27*26*25*24/(7*6*4)
    = 29*4*26*25 = 75,400

    then
    p = 75,400 (.2^7)(.8^23)
    = .005697

  • math -

    prob to go Disney = 1/5
    prob not to go Disney = 4/5

    prob that 7 of 30 will go
    = C(30,7)(1/5)^7 (4/5)^23 = appr. .1538

  • math -

    binomial coef error
    coef = 2035800 not 75,400
    30!/23!/7! = 30*29*28*27*26*25*24/(7*6*5*4*3*2)
    2035800

    p = 2035800 (.2^7)(.8^23)
    = .1538

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