SOLUTION: 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

Algebra ->  Probability-and-statistics -> SOLUTION: 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      Log On


   

Question 1160651: 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 ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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(a)  P = 1-0.86%29%5E3 = 0.14%5E3 = 0.002744.



(b)  P = 0.86%5E3 = 0.636056.



(c)  P = 1 - %281-0.086%29%5E3 = 1 - 0.002744 = 0.997256.

Solved.

All formulas are self-explanatory.