Respuesta :
Using the binomial distribution, it is found that there is a 0.99999984 = 99.999984% probability that at least one household has a television set.
For each household, there are only two possible outcomes, either it has at least one television set, or it does not. The probability of a household having at least one television set is independent of any other household, which means that the binomial distribution is used to solve this question.
Binomial probability distribution
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem:
- 4 households are selected, hence [tex]n = 4[/tex].
- 98% have at least one television set, hence [tex]p = 0.98[/tex]
The probability that at least one household has a television set is:
[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]
In which:
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{4,0}.(0.98)^{0}.(0.02)^{4} = 0.00000016[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.00000016 = 0.99999984 [/tex]
0.99999984 = 99.999984% probability that at least one household has a television set.
A similar problem is given at https://brainly.com/question/24863377
