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.
To find the probability that at least one of the three picked households will have cable television, we can use the principle of complementary probability. It states that the probability of an event happening is equal to one minus the probability of the event not happening.
To get started, let's find the probability that none of the three picked households have cable television. If 86% of households have cable television, then the probability that a randomly selected household does not have cable television is 1 - 0.86 = 0.14.
Now, since the households are picked at random, the probability of none of the three households having cable television is (0.14) * (0.14) * (0.14) = 0.002744.
Finally, we can subtract this probability from 1 to find the probability that at least one of the three households will have cable television: 1 - 0.002744 = 0.997256.
Therefore, the probability that at least one of the three picked households will have cable television is approximately 0.9973, or 99.73%.