.8 probability that a randomly selected adult will watch prime time tv live, instead of online, on DVR, etc.. Assume that seven adults are randomly selected. find the propbablity that fewer than three of the selected adults watch prime time live.

Answer 1

prob(watch live) = .8
prob(not live) = .2

fewer than three of 7 means: 0 live, 1 live, or 2 live

prob(of your stated event)
= C(7,0)(.8)^0 (.2)^7 + C(7,1)(.8)^1 (.2)^6 + C(7,2)(.8)^2 (.2)^5
= .0000128 + .0003584 + .0043008
= .004672

seems low since the distribution is skewed towards the other end
e.g. C(7,6)(.8)^6 (.2)^1 = .367 , that one entry alone accounts for over 1/3 of all the distribution.

Answer 2

this is a binomial probability ... prime or other ... p(P) = .8 ... p(O) = .2

(O + P)^8 = O^8 + 8 O^7 P + 28 O^6 P^2 + ... + P^8

sum the 1st three terms to find p(P<3)
... .2^8 + (8 * .2^7 * .8) + (28 * .2^6 * .8^2)

Answer 3

got the exponent wrong ... should be 7 , not 8

(O + P)^7 = O^7 + 7 O^6 P + 21 O^5 P^2 + ... + P^7

.2^7 + (7 * .2^6 * .8) + (21 * .2^5 * .8^2)