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.

Answer 1

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

Answer 2

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

Answer 3

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