Question

Suppose that in a large metropolitan area, 73% of all households have cable tv. Suppose you are interested in selecting a group of six households from this area. Let X be the number of households in a group of six households from this area that have cable tv. For what proportion of groups will at most two of the households have cable tv?

Answer 1

Answer:

$P(X \leq 2)=0.000387+0.00628+0.0425=0.0491$

Step-by-step explanation:

1) Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

2) Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=6, p=0.73)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

For what proportion of groups will at most two of the households have cable tv?

So we want to find this probability:

$P(X \leq 2)=P(X=0)+P(X=1)+P(X=2)$

$P(X=0)=(6C0)(0.73)^0 (1-0.73)^{6-0}=0.000387$

$P(X=1)=(6C1)(0.73)^1 (1-0.73)^{6-1}=0.00628$

$P(X=2)=(6C2)(0.73)^2 (1-0.73)^{6-2}=0.0425$

$P(X \leq 2)=0.000387+0.00628+0.0425=0.0491$