Question

A TV executive is interested in the popularity of a particular cable TV show. She has been told that a whopping 68% of American households would be interested in tuning in to a new network version of the show. If this is correct, what is the probability that all 6 of the households in her city being monitored by the TV industry would tune in to the new show

Answer 1

Answer:

0.0989 = 9.89% probability that all 6 of the households in her city being monitored by the TV industry would tune in to the new show

Step-by-step explanation:

For each household, there are only two possible outcomes. Either they will tune in to the show, or they wont. The probability of a household being tuned in to the show is independent of any other household. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

She has been told that a whopping 68% of American households would be interested in tuning in to a new network version of the show.

This means that $p = 0.68$

If this is correct, what is the probability that all 6 of the households in her city being monitored by the TV industry would tune in to the new show

All 6 is $P(X = 6)$ when $n = 6$. So

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{6,6}.(0.68)^{6}.(0.32)^{0} = 0.0989$

0.0989 = 9.89% probability that all 6 of the households in her city being monitored by the TV industry would tune in to the new show