What is the probability that at least one T.V. home in the sample is served by cable?

Question

New York City is estimated to have 21 percent of the homes with T.V. sets served by cable T.V. If a random sample of four is taken from these homes find the probability distribution of x where x is the number of T.V. homes in the sample having cable T.V.

What is the probability that at least one T.V. home in the sample is served by cable?

Answer 1

If the weatherman told you there was an 70% chance of rain tomorrow, you would know there was a 30% chance of NO RAIN, right?

Same logic applies here...

What is the probability that at least one TV home in the sample is served by cable?

Find the probability of NO CABLE in a sample of 4.

P(no cable) = (.79)^4 = .38950081 or 38.95%

So, the probability of getting AT LEAST one house with cable is:

1 - .38950081 = .61049919 or 61.05%

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To construct a probability distribution for x, use the Binomial Probability Formula:

P(N) = N! / [k! * (N - k)!] * p^k * q^(N - k)

where:

N = Number of opportunities for event x to occur (4)

k = Number of times that event x should happen (0)

p = Probability of a success (.21)

q = Probability of failure (.79)

! = Factorial

P(k) = N! / [k! * (N - k)!] * p^k * q^(N - k)

P(1 cable) = 4! / [1! * (4 - 1)!] * (.21)^1 * (.79)^(4 - 1)

P(1 cable) = .41415276 or 41.42%

P(no cable) =.38950081 or 38.95%

P(1 cable) = .41415276 or 41.42%

P(2 cable) = .16513686 or 16.51%

P(3 cable) = .02926476 or 2.93%

P(4 cable) = .00194481 or 0.19%

P(Total) = 1.00 or 100%

Yeah, the math gets really messy,

but you can use an online Binomial Probability Calculator:

http://faculty.vassar.edu/lowry/ch5apx.h...

Or you can do it on your TI-83/84 Calculator:

http://mathbits.com/MathBits/TISection/S...

Good luck in your studies,

~ Mitch ~