Posted by Janice on .
If 86% of households have cable television and three households are picked at random, what is the probability that at least one of those three picked will have cable television?
Use a binomial probability table or use the formula to do this by hand:
P(0) = (nCx)(p^x)[q^(n-x)]
n = 3
x = 0
p = .86
q = 1 - p
Substitute the values into the formula and calculate P(0). Then subtract from 1 for your probability.