Question

A TV​ show, Lindsay and Tobias​, recently had a share of 10​, meaning that among the TV sets in​ use, 10​% were tuned to that show. Assume that an advertiser wants to verify that 10​% share value by conducting its own​ survey, and a pilot survey begins with 8 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:

0.43 = 43% probability that none of the households are tuned to Lindsay and Tobias.

0.57 = 57% probability that at least one household is tuned to Lindsay and Tobias.

0.813 = 81.3% probability that at most one household is tuned to Lindsay and Tobias.

Step-by-step explanation:

For each household, there are only two possible outcomes. Either they are tuned to Lindsay and Tobias, or they are not. The probability of a household being tuned to Lindsay and Tobias is independent of other households. So we use the binomial probability distribution 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.

In this problem we have that:

$n = 8, p = 0.1$

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

This is $P(X = 0)$

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

$P(X = 0) = C_{8,0}.(0.1)^{0}.(0.9)^{8} = 0.43$

0.43 = 43% 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 none is tuned, or at least one is. The sum of the probabilities of these events is 100%. From the first question

p + 43 = 100

p = 57%

0.57 = 57% 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.

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

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

$P(X = 0) = C_{8,0}.(0.1)^{0}.(0.9)^{8} = 0.43$

$P(X = 0) = C_{8,1}.(0.1)^{1}.(0.9)^{7} = 0.383$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.43 + 0.383 = 0.813$

0.813 = 81.3% probability that at most one household is tuned to Lindsay and Tobias.