SOLUTION: Cable Television In 2006, 86% of U.S. households had
cable TV. Choose 3 households at random. Find the
probability that
a. None of the 3 households had cable TV
b. All 3 househ
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-> SOLUTION: Cable Television In 2006, 86% of U.S. households had
cable TV. Choose 3 households at random. Find the
probability that
a. None of the 3 households had cable TV
b. All 3 househ
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Question 376288: Cable Television In 2006, 86% of U.S. households had
cable TV. Choose 3 households at random. Find the
probability that
a. None of the 3 households had cable TV
b. All 3 households had cable TV
c. At least 1 of the 3 households had cable TV Answer by edjones(8007) (Show Source):
You can put this solution on YOUR website! Let x=.86, y=.14
(x+y)^3 Binomial expansion.
.
a) y^3=.0027
.
b) x^3=.6361
.
c)1-y^3=1-.0027=.9973
.
Ed
