Answer to Question #71800 in Statistics and Probability for Chris

Question

Having trouble with my homework. I can’t come up with the answers they get.
Q= Among 44-49 year olds 39% say they have danced in public while under the influence of alcohol. Suppose five 44-49 year olds are selected at random. (A) what is the probability that at least one has not danced in public under the influence of alcohol? (B) what is the probability that at least one has danced in public while under the influence of alcohol.

Answer 1

p = 0.39, q = 0.61, n = 5
Pn(k) = C(n;k)*p^k*q^(n - k),
C(n;k) = n!/(k!*(n - k)!).
(A) The probability that at least one has not danced in public under the influence of alcohol is:
P = 1 - C(5;0)*0.61^0*0.39^5 = 1 - 0.009 = 0.991 or 99.1%.
(B) The probability that at least one has danced in public while under the influence of alcohol is:
P = 1 - C(5;0)*0.39^0*0.61^5 = 1 - 0.085 = 0.915 or 91.5%.