Respuesta :
Using the binomial distribution, it is found that there is a 0.6521 = 65.21% probability that at least 1 is defective.
For each phone, there are only two possible outcomes, either it is defective, or it is not. The probability of a phone being defective is independent of any other phone, 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:
- 0.14 probability of a single phone being defective, hence [tex]p = 0.14[/tex]
- 7 phones are selected at random, hence [tex]n = 7[/tex]
The probability that at least 1 is defective 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_{7,0}.(0.14)^{0}.(0.86)^{7} = 0.3479[/tex]
Then:
[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.3479 = 0.6521[/tex]
0.6521 = 65.21% probability that at least 1 is defective.
A similar problem is given at https://brainly.com/question/24863377
