One Sample
What is the probability that at least one TV household in the sample is served by cable?
New York City is estimated at 21 percent of homes with televisions served by cable television If a random sample is drawn from four of these houses to find distribution probability of x where x is the number of TV households in the sample with cable TV What is the probability that at least one TV in the household sample is served by cable?
If the weather is telling you that there was a 70% chance of rain tomorrow, you know there was a reduction of 30% NO RAIN chance to right,? same logic applies here ... What is the probability that at least one TV household in the sample is served by cable? Find the probability of NO CABLE in a sample of 4. P (no cable) = (0.79) ^ 4 = 0.38950081 or 38.95% So, the probability of getting AT LEAST a house with cable is as follows: 1-0.38950081 = 0.61049919 or 61.05% --------------- -------- To build a distribution probability for x, use the binomial probability formula: P (N) = N! / [K! * (N - k)] k ^ * p * q ^ (n - k) where N = number of possibilities for the event to produce x (4) k = Number of times the event should occur x (0) p = probability of success (0.21) q = probability of failure (0.79)! P = factorial (k) = N! / [K! *! (N - k)] k ^ * p * q ^ (n - k) P (1 cable) = 4! / [1! *! (4-1)] * (.21) ^ 1 * (.79) ^ (4 - 1) P (1 cable) = 0.41415276 or 41.42% P (no cable) =. 38950081 or 38.95% (1 cable) P = 0.41415276 or 41.42% (2 wires) = 0.16513686 or 16.51% PP (3 cable) = .02926476, or 2.93% (4 cables) = 0.00194481 or 0.19% (Total) = 1.00 or 100% P Yeah, the calculation become really disorder, but you can use an online calculator binomial probability P: http://faculty.vassar.edu/lowry/ch5apx.html Or you can do on your TI-83/84 Calculator: Http://mathbits.com/MathBits/TISection/Statistics2/bernoulli.htm Good luck in your studies, ~ Mitch ~
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