Posted by **deborah** on Friday, July 22, 2011 at 8:10pm.

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 -
**Damon**, Friday, July 22, 2011 at 8:26pm
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 -
**Reiny**, Friday, July 22, 2011 at 8:26pm
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 -
**Damon**, Friday, July 22, 2011 at 9:00pm
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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