Question

A TV show, Lindsay and Tobias, recently had a share of 20, meaning that among the TV sets in use, 20% were tuned to that show. Assume that an advertiser wants to verify that 20% share value by conducting its own survey, and a pilot survey begins with 11 households having TV sets in use at the time of a Lindsay and Tobias broadcast.
Find the probability that none of the households are tuned to Lindsay and Tobias.

(Round to three decimal places as needed.)

Find the probability that at least one household is tuned to Lindsay and Tobias.

(Do not round until the final answer. Then round to three decimal places.)

Find the probability that at most one household is tuned to Lindsay and Tobias.

(Round to three decimal places as needed.)

Answer 1

Answer:

8.6% probability that none of the households are tuned to Lindsay and Tobias.

There is a 91.4% probability that at least one household is tuned to Lindsay and Tobias.

There is a 32.2% probability that at most one household is tuned to Lindsay and Tobias.

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they are watching the TV show, or they are not. This means that we use the binomial probability distribution to solve this problem.

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 combinatios 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.

In this problem we have that:

$n = 11, p = 0.20$

Find the probability that none of the households are tuned to Lindsay and Tobias.

This is P(X = 0).

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

$P(X = 0) = C_{11,0}.(0.20)^{0}.(0.80)^{11} = 0.086$

8.6% probability that none of the households are tuned to Lindsay and Tobias.

Find the probability that at least one household is tuned to Lindsay and Tobias.

Either no household is tuned, or at least one is tuned. The sum of the probabilities of these events is decimal 1. So

$P(X = 0) + P(X \geq 1) = 1$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.086 = 0.914$

There is a 91.4% probability that at least one household is tuned to Lindsay and Tobias.

Find the probability that at most one household is tuned to Lindsay and Tobias.

This is

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

$P(X = 1) = C_{11,1}.(0.20)^{1}.(0.80)^{10} = 0.236$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.086 + 0.236 = 0.322$

There is a 32.2% probability that at most one household is tuned to Lindsay and Tobias.