Question 279073
The cable TV industry reports that 59% of homes in the US to cable TV. If you pick four US households at random, what is the probability that at least one of the households is not a cable TV subscriber?
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Binomial Problem:
n = 4 ; p = 0.59 ; q = 0.41 ; 1<= x <=4
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P(at least one) = 1 - P(none)
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P(at least one = 1 - 4C0(0.59)^0*(0.41)^4 = 1-0.41^4 = 0.9717...
Cheers,
Stan H.